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

$$Foranycomplexnumberz,z^n=\overline{z}has(n+2)solutionsHow???$$

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

$$letz=r(\cos \theta +i\sin \theta )={re}^{i\theta }$$$$z^n=r^n(\cos n\theta +i\sin n\theta )$$$$\overline{z}=r(\cos \theta -i\sin \theta )$$$$r=0isasolution.$$$$forr\ne 0:$$$$r^n\cos n\theta =r\cos \theta \Rightarrow r^{n-1}=\frac{\cos \theta }{\cos n\theta }...\left(1\right)$$$$r^n\sin n\theta =-r\sin \theta \Rightarrow r^{n-1}=-\frac{\sin \theta }{\sin n\theta }...\left(2\right)$$$$\frac{\cos \theta }{\cos n\theta }=-\frac{\sin \theta }{\sin n\theta }$$$$\sin n\theta \cos \theta +\cos n\theta \sin \theta =1$$$$\sin (n+1)\theta =1$$$$\Rightarrow (n+1)\theta =2k\pi +\frac{\pi }{2}$$$$\Rightarrow \theta =\frac{2k\pi }{n+1}+\frac{\pi }{2(n+1)}withk=0,1,...,n$$$$i.e.therearen+1solutionsfor\theta$$$$andthereforeforrandtherefore$$$$forz.$$$$sototallytherearen+2solutions:$$$$z=0$$$$z_k=r_k(\cos {\theta }_k+i\sin {\theta }_k)=r_k{e}^{i{\theta }_k}$$$$with$$$${\theta }_k=\frac{2k\pi }{n+1}+\frac{\pi }{2(n+1)}$$$$r_k=\sqrt[n-1]{\frac{\cos {\theta }_k}{\cos n{\theta }_k}}$$$$andk=0,1,...,n$$

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