Explained: "If z is a complex function and z3=8i,..."

Pam Stokes

Answered question

2022-01-18

If z is a complex function and

z^{3} = 8 i

, what are all the values of z?

Vivian Soares

Beginner2022-01-19Added 36 answers

z^{3} = 8 i

z^{3} - 8 i = 0

z^{3} + {(2 i)}^{3} = 0

(z + 2 i) (z^{2} + {(2 i)}^{2} - 2 i z) = 0

(z + 2 i) (z^{2} - 2 i z - 4) = 0

z = - 2 i

z^{2} - 2 i z - 4 = 0

z = \frac{2 i \pm \sqrt{- 4 + 16}}{2}

z = i \pm \sqrt{3}

So roots are

- 2 i, i + \sqrt{3}, i - \sqrt{3}

aquariump9

Beginner2022-01-20Added 40 answers

z^{3} = 8 i = 8 e^{i (\frac{π}{2} + 2 n π)}

where

n \in Z

z = 2 e^{i (\frac{π}{6} + 2 n \frac{π}{3})}

z = 2 e^{i \frac{π}{6}}, 2 e^{i (5 \frac{π}{6})}, 2 e^{i (3 \frac{π}{2})}

z = \sqrt{3} + i, - \sqrt{3} + i, - 2 i

alenahelenash

Expert2022-01-24Added 556 answers

Let

ω = e^{\frac{2 i π}{3}} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}

is a root of unit. So the roots of

Z^{3} = 1 are : ω^{0} = 1; ω and ω^{2} = \overset{―}{ω}

We have:

z^{3} = 8 i \Leftrightarrow z^{3} = (- 2 i)^{3} \Leftrightarrow (\frac{z}{- 2 i})^{3} = 1

So:

\frac{z}{- 2 i} = ω^{k} k \in 0; 1; 2

. Thus The root of the equation

z^{3} = 8 i

are:

z = - 2 i {\dot{ω}}^{k} k \in 0; 1; 2 i . e . z = - 2 i or z = \sqrt{3} + i or z = - \sqrt{3} + i

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