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I don't know how I should solve this question:

If $0<\theta<\pi/2$ and $z=(\sin\theta+i(1-\cos\theta))^2$ find in its simplest form $\arg z$

I know that using de Moivre's theorem I should multipy angle by two but there is $i(1-\cosθ)$ and I don't know if I should multipy only $\cos\theta$ which would give $i-i\cos2\theta$ ?

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  • $\begingroup$ Just because something is called $\theta$, it does not mean that it is the angle you are looking for. De Moivre's formula says that there is exactly a real number $\alpha\in[0,2\pi)$ such that $\frac{z}{\sqrt{ z\overline z}}=\cos\alpha+i\sin\alpha$. For such real number $\alpha$, it is true that $z^2=(z\overline z)\left(\cos(2\alpha)+i\sin(2\alpha)\right)$. Doing the aforementioned calcs, it might turn out that $\alpha$ has a non-trivial dependence on $\theta$. $\endgroup$
    – user228113
    Commented May 29, 2017 at 17:05

1 Answer 1

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We have

\begin{align} z&=\left[2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+i\left(2\sin^2\frac{\theta}{2}\right)\right]^2\\ &=4\sin^2\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)^2\\ &=4\sin^2\frac{\theta}{2}\left(\cos\theta+i\sin\theta\right)\\ \end{align}

$\arg z=\theta$.

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  • $\begingroup$ I don't understand how did we get the second line. Could you explain me that, please? $\endgroup$
    – Markowska
    Commented May 29, 2017 at 17:19
  • $\begingroup$ $2\sin\frac{\theta}{2}$ is a common factor. So we have $\displaystyle 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+i\left(2\sin^2\frac{\theta}{2}\right) =2\sin\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)$ $\endgroup$
    – CY Aries
    Commented May 29, 2017 at 17:22
  • $\begingroup$ Oh, and you squared it at once. Now I can see it. Thank you very much! $\endgroup$
    – Markowska
    Commented May 29, 2017 at 17:26

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