I came across this question from an older textbook with no answers and I'm a bit stuck. Currently, I have done the following;

let $z=r(\cos x+i \sin x)$ ($z \in \mathbb{C}$)

therefore, we now have

$r^n(\cos nx + i \sin nx) = (r\cos x + i(r \sin x+1))^n$

From here its seems that I should equate the real and imaginary parts of the RHS with the real and imaginary parts of the LHS but after trying to do this, I'm not really getting anywhere. Also, $n$ cannot be $1$? So will I need to split it into cases of $n$? any help appreciated.

EDIT: Apologies for lack of workings; this is my first post so I'm not too sure what I'm doing. Upon reflection, I noticed the original title question was incorrect and I think I have changed it. Thank you to the users who have formatted the question for me. Also, I have NOT got the right answer. Here is where I am at now.

$z^n = (i+z)^n$

multiplying 'creatively' by one;

$\frac {z^n} {(i+z)^n} = 1 \\ \left( \frac z {i + z} \right) ^n = 1 \\ \frac z {z + i} = \cos \left( 2k \frac \pi n \right) + i \sin \left( 2k \frac \pi n \right)$

by De Moivre's Theorem, but now I am stuck. Any help appreciated.

  • 6
    $\begingroup$ Instead of deleting the question, post your own answer and mention that you figured it out. $\endgroup$ – Stella Biderman Feb 3 '16 at 15:49
  • 1
    $\begingroup$ If $z=r(\cos{\varphi}+i\sin{\varphi})$ then $1+z=(1+r\cos{\varphi})+ir\sin{\varphi}.$ $\endgroup$ – M. Strochyk Feb 3 '16 at 15:54


$(i+z)/z=e^{2k\pi i/n}$ where $0\le k<n$

$\dfrac iz=\cos\dfrac{2k\pi}n+i\sin\dfrac{2k\pi}n-1=2i\sin\dfrac{k\pi}n\left(\cos\dfrac{k\pi}n+i\sin\dfrac{k\pi}n\right)$

$2z=\dfrac{\cos\dfrac{k\pi}n-i\sin\dfrac{k\pi}n}{\sin\dfrac{k\pi}n}$ where $1\le k\le n-1$

  • $\begingroup$ Lab. $k=0$ is not permissible ($z+i\ne z$). Other values of $k$ not listed are permissible. $\endgroup$ – Mark Viola Feb 3 '16 at 20:12
  • $\begingroup$ how did you go from the second line to the third line? $\endgroup$ – gozzo98 Feb 3 '16 at 22:34
  • $\begingroup$ @Dr.MV, that is clear from the last line $\endgroup$ – lab bhattacharjee Feb 4 '16 at 1:43
  • $\begingroup$ Lab. The middle line ends with "$0\le 2k<n$," but it should be "$1\le k\le n-1$," should it not? ;-)) - Mark $\endgroup$ – Mark Viola Feb 4 '16 at 2:56
  • $\begingroup$ @Dr.MV, Agreed & rectified. Thanks $\endgroup$ – lab bhattacharjee Feb 4 '16 at 6:00

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