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Question Number 128459 by SLVR last updated on 07/Jan/21

F o r a n y c o m p l e x n u m b e r z, z^{n} = \bar{z} h a s (n + 2) s o l u t i o n s H o w ? ? ?

Answered by mr W last updated on 10/Jan/21

l e t z = r (\cos θ + i \sin θ) = {r e}^{i θ}

z^{n} = r^{n} (\cos n θ + i \sin n θ)

\bar{z} = r (\cos θ - i \sin θ)

r = 0 i s a s o l u t i o n .

f o r r \neq 0 :

r^{n} \cos n θ = r \cos θ \Rightarrow r^{n - 1} = \frac{\cos θ}{\cos n θ} \dots (1)

r^{n} \sin n θ = - r \sin θ \Rightarrow r^{n - 1} = - \frac{\sin θ}{\sin n θ} \dots (2)

\frac{\cos θ}{\cos n θ} = - \frac{\sin θ}{\sin n θ}

\sin n θ \cos θ + \cos n θ \sin θ = 1

\sin (n + 1) θ = 1

\Rightarrow (n + 1) θ = 2 k π + \frac{π}{2}

\Rightarrow θ = \frac{2 k π}{n + 1} + \frac{π}{2 (n + 1)} w i t h k = 0, 1, \dots, n

i . e . t h e r e a r e n + 1 s o l u t i o n s f o r θ

a n d t h e r e f o r e f o r r a n d t h e r e f o r e

f o r z .

s o t o t a l l y t h e r e a r e n + 2 s o l u t i o n s :

z = 0

z_{k} = r_{k} (\cos θ_{k} + i \sin θ_{k}) = r_{k} e^{i θ_{k}}

w i t h

θ_{k} = \frac{2 k π}{n + 1} + \frac{π}{2 (n + 1)}

r_{k} = \sqrt[n - 1]{\frac{\cos θ_{k}}{\cos n θ_{k}}}

a n d k = 0, 1, \dots, n

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