Step by step solution to "How do I solve z5=−2i (complex numbers)?"

sunshine022uv

Answered question

2022-01-16

How do I solve

z^{5} = - 2 i

(complex numbers)?

sirpsta3u

Beginner2022-01-17Added 42 answers

z^{5} = - 2 i

Convert the right side into exponential form

z^{5} = 2 e^{- \frac{i π}{2} + 2 k i π}

z = \sqrt[5]{2} e^{- \frac{π}{10} + \frac{2 k i π}{5}}

z = \sqrt[5]{2} [\cos (4 k + 1) \frac{π}{10} + i \sin (4 k + 1) \frac{π}{10}]

Putting

k = 0, 1, 2, 3, 4. . .

z_{1} = \sqrt[5]{2} [\cos \frac{π}{10} + i \sin \frac{π}{10}]

z_{2} = \sqrt[5]{2} [\cos \frac{π}{2} + i \sin \frac{π}{2}] = \sqrt[5]{2} i

z_{3} = \sqrt[5]{2} [\cos \frac{9 π}{10} + i \sin \frac{9 π}{10}]

z_{4} = \sqrt[5]{2} [\cos \frac{π}{10} - i \sin \frac{π}{10}]

z_{5} = \sqrt[5]{2} [\cos \frac{9 π}{10} - i \sin \frac{9 π}{10}]

levurdondishav4

Beginner2022-01-18Added 38 answers

I would use polar coordinates, as extracting roots becomes far easier.

- 2 i = 2 e^{\frac{3 π}{2}}

z = [2^{\frac{1}{5}}] \cdot e^{3 \frac{π}{10} + 2 \cdot j \cdot \frac{π}{5}}

where

j = 0, 1, 2, 3, or 4

to get the 5 roots of this equation.
I used the fact that

{[e^{i \cdot x}]}^{\frac{1}{5}} = e^{\frac{i x}{5}}

. This comes from the identity for exponents that

{(e^{a})}^{b} = e^{a b}

alenahelenash

Expert2022-01-24Added 556 answers

\Rightarrow - 2 i = 2 (e^{- i p / 2})

\Rightarrow therefore z = 2^{0.2} (e^{- i p / 2})^{\frac{1}{5}} where, p = π e and i = \sqrt{- 1}

; now by applying de morvies

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