Solve-z-4-16-

Question Number 12442 by tawa last updated on 22/Apr/17

Solve : z^{4} = - 16

Answered by ajfour last updated on 22/Apr/17

z^{4} = 2^{4} e^{i (π + 2 k π)}

z = 2 e^{i (\frac{π}{4} + \frac{2 k π}{4})} \forall k = 0, 1, - 1, - 2

s o z_{1} = 2 [\cos (π / 4) + i \sin (π / 4)]

z_{2} = 2 [\cos (3 π / 4) + i \sin (3 π / 4)]

z_{3} = 2 [\cos (- π / 4) + i \sin (- π / 4)]

z_{4} = 2 [\cos (- 3 π / 4) + i \sin (- 3 π / 4)]

t o s u m m a r i z e

z = \sqrt{2} (\pm 1 \pm i) .

Commented by tawa last updated on 22/Apr/17

I really appreciate sir . God bless you .

Answered by mrW1 last updated on 23/Apr/17

l e t z = r (\cos θ + i \sin θ)

z^{4} = r^{4} (\cos 4 θ + i \sin 4 θ) = - 16

r^{4} = 16 \Rightarrow r = 2

\sin 4 θ = 0

\cos 4 θ = - 1

\Rightarrow 4 θ = (2 n - 1) π

\Rightarrow θ = n \frac{π}{2} - \frac{π}{4}

\cos θ = \pm \frac{\sqrt{2}}{2}

\sin θ = \pm \frac{\sqrt{2}}{2}

\Rightarrow z = 2 (\pm \frac{\sqrt{2}}{2} \pm i \frac{\sqrt{2}}{2}) = \sqrt{2} (\pm 1 \pm 1 i)

Commented by tawa last updated on 23/Apr/17

I appreciate sir . God bless you sir .

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