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Lekcja 5 – Pierwiastkowanie

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Przeglądasz 11 wpisów - od 1 do 11 (z 11 łącznie)
  • Ten temat ma 10 odpowiedzi, 5 udzielających się, ostatnio wpisał/a coś 1 rok, 2 miesiące temu Joanna Grochowska.
  • Autor
    Posty
  • #10407
    Krystian Karczyński
    Dyrektor

      Pobierz wzory („trzy tabelki”) (PDF) Pobierz Lekcję na twardy dysk
    [Zobacz cały post na stronie: Lekcja 5 – Pierwiastkowanie (VIDEO)]

    #11820
    Anonim
    Nieaktywne

    witam
    panie Krystianie nie wychodza mi dobre wyniki w rozwiazaniu zadania domowego dotyczacego pierwiastkowania wychodza mi inne znaki

    #11819
    Krystian Karczyński
    Dyrektor

    Witam

    A o który konkretnie przykład chodzi? Chętnie pomogę.

    #11222
    Krystian Karczyński
    Dyrektor

    13.02.2015 KOREKTA

    Odpowiedź do Zadania 6 z część praktycznej.

    Poprawiłem odpowiedź na prawidłową, tzn:

     \left\{ \begin{matrix} & 1+i \\ & -\frac{1}{2}-\frac{\sqrt{3}}{2}+i( -\frac{1}{2}+\frac{\sqrt{3}}{2} ) \\ & -\frac{1}{2}+\frac{\sqrt{3}}{2}+i( -\frac{1}{2}-\frac{\sqrt{3}}{2} ) \\ \end{matrix} \right.

    • Ta odpowiedź została zmodyfikowana 1 rok, 7 miesiące temu przez Krystian Karczyński.
    #10764
    Krystian Karczyński
    Dyrektor

    19.10.2015 ERRATA

    Zadanie 6

    Pierwiastki wyliczone są prawidłowo, jednak 2 ostatnie nie można przedstawić w postaci trygonometrycznej bez zaawansowanych tabli, powinna więc być odpowiedź:

    cube root of open parentheses 1 plus i close parentheses cubed end root equals open curly brackets table row cell 1 plus i end cell row cell square root of 2 open parentheses cos 11 over 12 straight \pi plus isin 11 over 12 straight \pi close parentheses end cell row cell square root of 2 open parentheses c o s 19 over 12 straight \pi plus i s i n 19 over 12 straight \pi close parentheses end cell end table close

    #10763
    Krystian Karczyński
    Dyrektor

    Zadanie 6

    W każdym razie, podaję ten przykład krok po kroku, „sposobem 1” (uwaga: szybciej by poszło korzystając ze „sposobu 2”):

    cube root of open parentheses 1 plus i close parentheses cubed end root equals cube root of 1 cubed plus 3 \times 1 \times i squared plus 3 \times 1 squared \times i plus i cubed end root equals cube root of 1 minus 3 plus 3 i minus i end root equals cube root of negative 2 plus 2 i end root

    Przekształcam liczbę pod pierwiastkiem, tzn. z equals negative 2 plus 2 i na postać trygonometryczną, korzystając z „trzech tabelek”:

    „Trzy tabelki” do przechodzenia na postać trygonometryczną

    open vertical bar z close vertical bar equals square root of open parentheses negative 2 close parentheses squared plus 2 squared end root equals square root of 4 plus 4 end root equals square root of 8 equals square root of 4 \times 2 end root equals 2 square root of 2 cos \phi equals \fraction numerator negative 2 over denominator 2 square root of 2 end \fraction equals negative \fraction numerator 1 over denominator square root of 2 end \fraction equals negative \fraction numerator square root of 2 over denominator 2 end \fraction sin \phi equals \fraction numerator 2 over denominator 2 square root of 2 end \fraction equals \fraction numerator 1 over denominator square root of 2 end \fraction equals \fraction numerator square root of 2 over denominator 2 end \fraction

    Ze względu na znaki, orientuję się, że argument główny będzie w II ćwiartce, więc wyrazić go można jako:

    \phi equals \pi minus \alpha

    Patrząc na wartości cos \phi i sin \phi odczytuję z odpowiedniej tabelki, że \alpha equals \pi over 4, więc mam:

    \phi equals \pi minus \alpha equals \pi minus \pi over 4 equals 3 over 4 pi

    Cała więc liczba z equals negative 2 plus 2 i w postaci trygonometrycznej będzie równa:

    negative 2 plus 2 i equals 2 square root of 2 open parentheses cos 3 over 4 straight \pi plus isin 3 over 4 straight \pi close parentheses

    Mogę przystąpić do liczenia jej trzech pierwiastków, korzystając z wzoru podanego na

    Lekcja 5 – Pierwiastki

    omega subscript 0 equals cube root of 2 square root of 2 end root open parentheses cos \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 0 \times straight \pi end sty\le over denominator 3 end \fraction plus i sin \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 0 \times straight \pi end sty\le over denominator 3 end \fraction close parentheses equals cube root of square root of 8 end root open parentheses c o s \fraction numerator begin display sty\le 3 over 4 straight \pi end sty\le over denominator 3 end \fraction plus i s i n \fraction numerator begin display sty\le 3 over 4 straight \pi end sty\le over denominator 3 end \fraction close parentheses equals equals square root of cube root of 8 end root open parentheses c o s open parentheses 3 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses plus i s i n open parentheses 3 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses close parentheses equals square root of 2 open parentheses c o s 1 fourth straight \pi plus i s i n 1 fourth straight \pi close parentheses equals square root of 2 open parentheses \fraction numerator square root of 2 over denominator 2 end \fraction plus i \fraction numerator square root of 2 over denominator 2 end \fraction close parentheses equals equals 2 over 2 plus i 2 over 2 equals 1 plus i

    omega subscript 1 equals cube root of 2 square root of 2 end root open parentheses cos \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 1 \times straight \pi end sty\le over denominator 3 end \fraction plus i sin \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 1 \times straight \pi end sty\le over denominator 3 end \fraction close parentheses equals cube root of square root of 8 end root open parentheses c o s \fraction numerator begin display sty\le 2 3 over 4 straight \pi end sty\le over denominator 3 end \fraction plus i s i n \fraction numerator begin display sty\le 2 3 over 4 straight \pi end sty\le over denominator 3 end \fraction close parentheses equals equals square root of cube root of 8 end root open parentheses c o s open parentheses 11 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses plus i s i n open parentheses 11 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses close parentheses equals square root of 2 open parentheses c o s 11 over 12 straight \pi plus i s i n 11 over 12 straight \pi close parentheses

    omega subscript 2 equals cube root of 2 square root of 2 end root open parentheses cos \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 2 \times straight \pi end sty\le over denominator 3 end \fraction plus i sin \fraction numerator begin display sty\le 3 over 4 straight \pi plus 2 \times 2 \times straight \pi end sty\le over denominator 3 end \fraction close parentheses equals cube root of square root of 8 end root open parentheses c o s \fraction numerator begin display sty\le 4 3 over 4 straight \pi end sty\le over denominator 3 end \fraction plus i s i n \fraction numerator begin display sty\le 4 3 over 4 straight \pi end sty\le over denominator 3 end \fraction close parentheses equals equals square root of cube root of 8 end root open parentheses c o s open parentheses 19 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses plus i s i n open parentheses 19 over 4 straight \pi \times \fraction numerator begin display sty\le 1 end sty\le over denominator 3 end \fraction close parentheses close parentheses equals square root of 2 open parentheses c o s 19 over 12 straight \pi plus i s i n 19 over 12 straight \pi close parentheses

    Mamy więc odpowiedź:

    cube root of open parentheses 1 plus i close parentheses cubed end root equals open curly brackets table row cell 1 plus i end cell row cell square root of 2 open parentheses c o s 11 over 12 straight \pi plus i s i n 11 over 12 straight \pi close parentheses end cell row cell square root of 2 open parentheses c o s 19 over 12 straight \pi plus i s i n 19 over 12 straight \pi close parentheses end cell end table close

    #10410
    Anonim
    Nieaktywne

    A jak zrobić zadanie typu: z=fourth root of \fraction numerator negative 18 over denominator i square root of 3 plus 1 end \fraction end root ?

    #10408
    Joanna Grochowska
    Dyrektor

    fourth root of \fraction numerator negative 18 over denominator i square root of 3 plus 1 end \fraction end root

    Rozpiszę wyrażenie pod pierwiastkiem usuwając „niewymierność” z mianownika – mnożę przez sprzężenie:

    \fraction numerator negative 18 over denominator i square root of 3 plus 1 end \fraction \times \fraction numerator i square root of 3 minus 1 over denominator i square root of 3 minus 1 end \fraction equals \fraction numerator negative 18 open parentheses i square root of 3 minus 1 close parentheses over denominator i squared \times 3 minus 1 end \fraction equals \fraction numerator negative 18 open parentheses i square root of 3 minus 1 close parentheses over denominator negative 3 minus 1 end \fraction equals \fraction numerator negative 18 open parentheses i square root of 3 minus 1 close parentheses over denominator negative 4 end \fraction equals 9 over 2 open parentheses i square root of 3 minus 1 close parentheses equals 9 over 2 square root of 3 i minus 9 over 2

    Chcę więc wyznaczyć pierwiastki liczby zespolonejfourth root of 9 over 2 square root of 3 i minus 9 over 2 end root . Przekształcam liczbę pod pierwiastkiem, tzn. z equals negative 9 over 2 plus 9 over 2 square root of 3 i   na postać trygonometryczną, korzystając z „trzech tabelek”:

    „Trzy tabelki” do przechodzenia na postać trygonometryczną

     

    open vertical bar z close vertical bar equals square root of open parentheses negative 9 over 2 close parentheses squared plus open parentheses \fraction numerator 9 square root of 3 over denominator 2 end \fraction close parentheses squared end root equals square root of 81 over 4 plus \fraction numerator 81 \times 3 over denominator 4 end \fraction end root equals square root of 324 over 4 end root equals square root of 81 equals 9cos \phi equals \fraction numerator x over denominator open vertical bar z close vertical bar end \fraction equals \fraction numerator negative begin display sty\le 9 over 2 end sty\le over denominator 9 end \fraction equals negative 9 over 2 \times 1 over 9 equals negative 1 half
    sin \phi equals \fraction numerator y over denominator open vertical bar z close vertical bar end \fraction equals \fraction numerator begin display sty\le \fraction numerator 9 square root of 3 over denominator 2 end \fraction end sty\le over denominator 9 end \fraction equals \fraction numerator square root of 3 over denominator 2 end \fraction

    Patrząc na znaki w pierwszej tabelce, orientuję się, że argument główny będzie w II ćwiartce, więc wyrazić go można jako:
    \phi equals straight \pi minus \alpha subscript 0

    Patrząc na wartości cos \phi  i  sin \phi  odczytuję z trzeciej tabelki, że \alpha subscript 0 equals straight \pi over 3, więc mam:
    \phi equals straight \pi minus \alpha subscript 0 equals straight \pi minus straight \pi over 3 equals \fraction numerator 2 straight \pi over denominator 3 end \fraction

    Cała więc liczba z equals negative 9 over 2 plus 9 over 2 square root of 3 i  w postaci trygonometrycznej będzie równa:

    negative 9 over 2 plus 9 over 2 square root of 3 i equals 9 open parentheses cos 2 over 3 straight \pi plus straight i space sin 2 over 3 straight \pi close parentheses

    Przechodzę do liczenia pierwiastków, zgodnie z wzorem z Lekcji:

    n-th root of z equals n-th root of open vertical bar z close vertical bar end root space open parentheses cos open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses plus i space sin open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses close parentheses

    Mam więc:

    bold italic omega subscript bold 0 equals fourth root of 9 open parentheses cos open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 0 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses plus i space sin open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 0 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses close parentheses equals square root of 3 open parentheses cos open parentheses 2 over 3 straight \pi \times 1 fourth close parentheses plus i space sin open parentheses 2 over 3 straight \pi \times 1 fourth close parentheses close parentheses equals
    square root of 3 open parentheses cos open parentheses 1 over 6 straight \pi close parentheses plus straight i space sin open parentheses 1 over 6 straight \pi close parentheses close parentheses equals square root of 3 open parentheses \fraction numerator square root of 3 over denominator 2 end \fraction plus i space 1 half close parentheses equals 3 over 2 plus i space \fraction numerator square root of 3 over denominator 2 end \fraction

    bold italic omega subscript bold 1 equals fourth root of 9 open parentheses cos open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 1 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses plus i space sin open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 1 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses close parentheses equals square root of 3 open parentheses cos open parentheses 8 over 3 straight \pi \times 1 fourth close parentheses plus i space sin open parentheses 8 over 3 straight \pi \times 1 fourth close parentheses close parentheses equals
    square root of 3 open parentheses c o s open parentheses 2 over 3 straight \pi close parentheses plus straight i space s i n open parentheses 2 over 3 straight \pi close parentheses close parentheses equals square root of 3 open parentheses c o s open parentheses straight \pi minus 1 third straight \pi close parentheses plus straight i space s i n open parentheses straight \pi minus 1 third straight \pi close parentheses close parentheses equals
    square root of 3 open parentheses negative c o s open parentheses 1 third straight \pi close parentheses plus straight i space s i n open parentheses 1 third straight \pi close parentheses close parentheses equals square root of 3 open parentheses negative 1 half plus i space \fraction numerator square root of 3 over denominator 2 end \fraction close parentheses equals negative \fraction numerator square root of 3 over denominator 2 end \fraction plus i space 3 over 2

    bold italic omega subscript bold 2 equals fourth root of 9 open parentheses cos open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 2 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses plus i space sin open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 2 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses close parentheses equals square root of 3 open parentheses cos open parentheses 14 over 3 straight \pi \times 1 fourth close parentheses plus i space sin open parentheses 14 over 3 straight \pi \times 1 fourth close parentheses close parentheses equals
    square root of 3 open parentheses c o s open parentheses 7 over 6 straight \pi close parentheses plus straight i space s i n open parentheses 7 over 6 straight \pi close parentheses close parentheses equals square root of 3 open parentheses c o s open parentheses straight \pi plus 1 over 6 straight \pi close parentheses plus straight i space s i n open parentheses straight \pi plus 1 over 6 straight \pi close parentheses close parentheses equals
    square root of 3 open parentheses negative c o s open parentheses 1 over 6 straight \pi close parentheses minus straight i space s i n open parentheses 1 over 6 straight \pi close parentheses close parentheses equals square root of 3 open parentheses negative \fraction numerator square root of 3 over denominator 2 end \fraction minus i space 1 half close parentheses equals negative 3 over 2 minus i space \fraction numerator square root of 3 over denominator 2 end \fraction

    bold italic omega subscript bold 3 equals fourth root of 9 open parentheses cos open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 3 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses plus i space sin open parentheses \fraction numerator begin display sty\le 2 over 3 straight \pi plus 2 \times 3 \times straight \pi end sty\le over denominator 4 end \fraction close parentheses close parentheses equals square root of 3 open parentheses cos open parentheses 20 over 3 straight \pi \times 1 fourth close parentheses plus i space sin open parentheses 20 over 3 straight \pi \times 1 fourth close parentheses close parentheses equals
    square root of 3 open parentheses c o s open parentheses 5 over 3 straight \pi close parentheses plus straight i space s i n open parentheses 5 over 3 straight \pi close parentheses close parentheses equals square root of 3 open parentheses c o s open parentheses 2 straight \pi minus 1 third straight \pi close parentheses plus straight i space s i n open parentheses 2 straight \pi minus 1 third straight \pi close parentheses close parentheses equals
    square root of 3 open parentheses c o s open parentheses 1 third straight \pi close parentheses minus straight i space s i n open parentheses 1 third straight \pi close parentheses close parentheses equals square root of 3 open parentheses 1 half minus i space \fraction numerator square root of 3 over denominator 2 end \fraction close parentheses equals \fraction numerator square root of 3 over denominator 2 end \fraction minus i space 3 over 2

    Czyli odpowiedź:

    z subscript 1 equals 3 over 2 plus i space \fraction numerator square root of 3 over denominator 2 end \fraction
    z subscript 2 equals negative \fraction numerator square root of 3 over denominator 2 end \fraction plus i space 3 over 2
    z subscript 3 equals negative 3 over 2 minus i space \fraction numerator square root of 3 over denominator 2 end \fraction
    z subscript 4 equals \fraction numerator square root of 3 over denominator 2 end \fraction minus i space 3 over 2

    #68972
    Jakub Myrcik
    Student

    mi również w zadaniu domowym przykładzie 2 i 3 wychodzą inne znaki tzn w obu przykładach 3 i 4 pierwiastek wychodzą jakby na odwrotnie do odpowiedzi czyli zamiast w 3 pierwiastku ma byc sqrt 2 / 2 – sqrt 2/2 to mi to wychodzi w 4 pierwiastku

    #70300
    Joanna Grochowska
    Dyrektor

    Panie Jakubie rozumiem, że ma Pan identyczne wszystkie 4 pierwiastki? Tylko że są jakby w innej kolejności? Trzeci z czwartym zamienione? Jeśli tak to nie problem 🙂 Najważniejsze to by ostateczne końcowe wyniki się zgadzały.

    Mimo wszystko rozpiszę oba zadania.

    Na to wychodzi, że Panu dobrze wyszło, Krystian po prostu odpowiedzi napisał w „innej kolejności”. Najważniejsze by wszystkie wyniki się zgadzały 🙂

     

    Zadanie 2

    fourth root of negative 1 end root space equals space ?

     

    Przekształcam liczbę pod pierwiastkiem, tzn. z equals negative 1 space equals space minus 1 plus 0 i  na postać trygonometryczną, korzystając z „trzech tabelek”:

    „Trzy tabelki” do przechodzenia na postać trygonometryczną

     

    open vertical bar z close vertical bar equals square root of left parenthesis negative 1 right parenthesis squared plus 0 squared end root equals square root of 1 equals 1
cos phi equals fraction numerator x over denominator open vertical bar z close vertical bar end fraction equals fraction numerator negative 1 over denominator 1 end fraction equals negative 1
sin phi equals fraction numerator y over denominator open vertical bar z close vertical bar end fraction equals 0 over 1 equals 0

     

    Patrząc na znaki w pierwszej tabelce, orientuję się, że cosinus ujemy oraz sinus dodatni (przyjmuję podobnie jak Krystian w lekcji zero za „dodatnią” liczbę), więc argument główny będzie w II ćwiartce, stąd wyrazić go można jako: phi equals pi minus alpha subscript 0 .

    Patrząc na wartości  cos phi  i   sin phi  odczytuję z trzeciej tabelki, że alpha subscript 0 equals 0 degree , więc mam: phi equals pi minus alpha subscript 0 equals straight pi minus 0 equals bold pi

    Cała więc liczba z equals negative 1 space equals space minus 1 plus 0 i   w postaci trygonometrycznej będzie równa:

    negative 1 space equals 1 open parentheses cosπ space plus space i sinπ close parentheses

    Przechodzę do liczenia pierwiastków, zgodnie z wzorem z Lekcji:

    n-th root of z equals n-th root of open vertical bar z close vertical bar end root space open parentheses cos open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses plus i space sin open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses close parentheses

    Mam więc:

    bold italic omega subscript bold 0 equals fourth root of 1 open parentheses cos open parentheses fraction numerator straight pi plus 2 times 0 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator straight pi plus 2 times 0 times straight pi over denominator 4 end fraction close parentheses close parentheses equals 1 open parentheses cos straight pi over 4 plus i space sin straight pi over 4 close parentheses space equals
space space space space space space space equals bold space fraction numerator square root of bold 2 over denominator bold 2 end fraction bold plus bold italic i bold space fraction numerator square root of bold 2 over denominator bold 2 end fraction

    bold italic omega subscript bold 1 equals fourth root of 1 open parentheses cos open parentheses fraction numerator straight pi plus 2 times bold 1 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator straight pi plus 2 times bold 1 times straight pi over denominator 4 end fraction close parentheses close parentheses equals 1 open parentheses cos fraction numerator 3 straight pi over denominator 4 end fraction plus i space sin fraction numerator 3 straight pi over denominator 4 end fraction close parentheses space equals
equals space cos stack stack open parentheses straight pi minus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses with underbrace below with bold II bold space bold ćwiartka below plus i space sin open parentheses straight pi minus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses equals bold minus cos straight pi over 4 plus i space sin straight pi over 4 equals bold minus fraction numerator square root of bold 2 over denominator bold 2 end fraction bold plus bold italic i bold space fraction numerator square root of bold 2 over denominator bold 2 end fraction

    bold italic omega subscript bold 2 equals fourth root of 1 open parentheses cos open parentheses fraction numerator straight pi plus 2 times bold 2 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator straight pi plus 2 times bold 2 times straight pi over denominator 4 end fraction close parentheses close parentheses equals 1 open parentheses cos fraction numerator 5 straight pi over denominator 4 end fraction plus i space sin fraction numerator 5 straight pi over denominator 4 end fraction close parentheses space equals
equals space cos stack stack open parentheses straight pi plus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses with underbrace below with bold III bold space bold ćwiartka below plus i space sin open parentheses straight pi plus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses equals bold minus cos straight pi over 4 bold minus i space sin straight pi over 4 equals bold minus fraction numerator square root of bold 2 over denominator bold 2 end fraction bold minus bold italic i bold space fraction numerator square root of bold 2 over denominator bold 2 end fraction

    bold italic omega subscript bold 3 equals fourth root of 1 open parentheses cos open parentheses fraction numerator straight pi plus 2 times bold 3 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator straight pi plus 2 times bold 3 times straight pi over denominator 4 end fraction close parentheses close parentheses equals 1 open parentheses cos fraction numerator 7 straight pi over denominator 4 end fraction plus i space sin fraction numerator 7 straight pi over denominator 4 end fraction close parentheses space equals
equals space cos stack stack open parentheses 2 straight pi minus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses with underbrace below with bold IV bold space bold ćwiartka below plus i space sin open parentheses 2 straight pi minus fraction numerator 1 straight pi over denominator 4 end fraction close parentheses equals cos straight pi over 4 bold minus i space sin straight pi over 4 equals fraction numerator square root of bold 2 over denominator bold 2 end fraction bold minus bold italic i bold space fraction numerator square root of bold 2 over denominator bold 2 end fraction

    #70310
    Joanna Grochowska
    Dyrektor

    Zadanie 3

    fourth root of negative 2 minus 2 square root of 3 i end root space equals space ?

    Przekształcam liczbę pod pierwiastkiem, tzn. z equals negative 2 minus 2 square root of 3 i  na postać trygonometryczną, korzystając z „trzech tabelek”:

    „Trzy tabelki” do przechodzenia na postać trygonometryczną

     

    z equals space minus 2 minus 2 square root of 3 i
open vertical bar z close vertical bar equals square root of left parenthesis negative 2 right parenthesis squared plus open parentheses negative 2 square root of 3 close parentheses squared end root equals square root of 4 plus 12 end root equals square root of 16 equals 4
cos phi equals fraction numerator x over denominator open vertical bar z close vertical bar end fraction equals fraction numerator negative 2 over denominator 4 end fraction equals negative 1 half
sin phi equals fraction numerator y over denominator open vertical bar z close vertical bar end fraction equals fraction numerator negative 2 square root of 3 over denominator 4 end fraction equals negative fraction numerator square root of 3 over denominator 2 end fraction

    Patrząc na znaki w pierwszej tabelce (cosinus ujemy oraz sinus ujemny), widzę, że argument główny będzie w III ćwiartce, stąd wyrazić go można jako: phi equals pi plus alpha subscript 0 .

    Patrząc na wartości  cos phi  i   sin phi  odczytuję z trzeciej tabelki, że alpha subscript 0 equals straight pi over 3 , więc mam: phi equals pi plus alpha subscript 0 equals straight pi plus straight pi over 3 equals fraction numerator bold 4 bold pi over denominator bold 3 end fraction

    Cała więc liczba z equals negative 2 minus 2 square root of 3 i   w postaci trygonometrycznej będzie równa:

    negative 2 minus 2 square root of 3 i space equals space 4 open parentheses cos fraction numerator 4 pi over denominator 3 end fraction space plus space i sin fraction numerator 4 pi over denominator 3 end fraction close parentheses

    Przechodzę do liczenia pierwiastków, zgodnie z wzorem z Lekcji:

    n-th root of z equals n-th root of open vertical bar z close vertical bar end root space open parentheses cos open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses plus i space sin open parentheses \fraction numerator \phi plus 2 k \pi over denominator n end \fraction close parentheses close parentheses

    Mam więc:

    bold italic omega subscript bold 0 equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times 0 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times 0 times straight pi over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos fraction numerator 4 straight pi over denominator 12 end fraction plus i space sin fraction numerator 4 straight pi over denominator 12 end fraction close parentheses space
space space space space space space space equals bold space fourth root of 4 open parentheses c o s straight pi over 3 plus i space s i n straight pi over 3 close parentheses bold space bold equals root index bold 4 of bold 4 bold space open parentheses bold 1 over bold 2 bold plus bold i bold space fraction numerator square root of bold 3 over denominator bold 2 end fraction close parentheses

    bold italic omega subscript bold 1 equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 1 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 1 times straight pi over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 straight pi over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 6 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 6 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 10 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 10 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos fraction numerator 10 straight pi over denominator 12 end fraction plus i space sin fraction numerator 10 straight pi over denominator 12 end fraction close parentheses space equals fourth root of 4 open parentheses cos fraction numerator 5 straight pi over denominator 6 end fraction plus i space sin fraction numerator 5 straight pi over denominator 6 end fraction close parentheses space equals
space space space equals fourth root of 4 space open parentheses cos stack stack open parentheses straight pi minus fraction numerator 1 straight pi over denominator 6 end fraction close parentheses with underbrace below with bold II bold space bold ćwiartka below plus i space sin open parentheses straight pi minus fraction numerator 1 straight pi over denominator 6 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses bold minus cos straight pi over 6 plus straight i space sin straight pi over 6 close parentheses equals root index bold 4 of bold 4 open parentheses bold minus fraction numerator square root of bold 3 over denominator bold 2 end fraction bold plus bold 1 over bold 2 bold i bold space close parentheses

    bold italic omega subscript bold 2 equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 2 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 2 times straight pi over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 4 straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 4 straight pi over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 12 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 12 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 16 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 16 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos fraction numerator 16 straight pi over denominator 12 end fraction plus i space sin fraction numerator 16 straight pi over denominator 12 end fraction close parentheses space equals fourth root of 4 open parentheses cos fraction numerator 4 straight pi over denominator 3 end fraction plus i space sin fraction numerator 4 straight pi over denominator 3 end fraction close parentheses space equals
space space space equals fourth root of 4 space open parentheses cos stack stack open parentheses straight pi plus fraction numerator 1 straight pi over denominator 3 end fraction close parentheses with underbrace below with bold III bold space bold ćwiartka below plus i space sin open parentheses straight pi plus fraction numerator 1 straight pi over denominator 3 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses bold minus cos straight pi over 3 bold minus straight i space sin straight pi over 3 close parentheses equals root index bold 4 of bold 4 open parentheses bold minus bold 1 over bold 2 bold minus fraction numerator square root of bold 3 over denominator bold 2 end fraction bold i bold space close parentheses

    bold italic omega subscript bold 3 equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 3 times straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 2 times bold 3 times straight pi over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 6 straight pi over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus 6 straight pi over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 18 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 4 straight pi over denominator 3 end fraction plus fraction numerator 18 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos open parentheses fraction numerator fraction numerator 22 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses plus i space sin open parentheses fraction numerator fraction numerator 22 straight pi over denominator 3 end fraction over denominator 4 end fraction close parentheses close parentheses
space space space equals fourth root of 4 open parentheses cos fraction numerator 22 straight pi over denominator 12 end fraction plus i space sin fraction numerator 22 straight pi over denominator 12 end fraction close parentheses space equals fourth root of 4 open parentheses cos fraction numerator 11 straight pi over denominator 6 end fraction plus i space sin fraction numerator 11 straight pi over denominator 6 end fraction close parentheses space equals
space space space equals fourth root of 4 space open parentheses cos stack stack open parentheses 2 straight pi minus fraction numerator 1 straight pi over denominator 6 end fraction close parentheses with underbrace below with bold IV bold space bold ćwiartka below plus i space sin open parentheses 2 straight pi minus fraction numerator 1 straight pi over denominator 6 end fraction close parentheses close parentheses equals fourth root of 4 open parentheses cos straight pi over 6 bold minus straight i space sin straight pi over 6 close parentheses equals root index bold 4 of bold 4 open parentheses fraction numerator square root of bold 3 over denominator bold 2 end fraction bold minus bold 1 over bold 2 bold i bold space close parentheses

     

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