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$\text { Show that }\left|z+z^{2}\right|=2 \cos \frac{\theta}{2}$

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It is known that $OABC$ and $z=cis(\theta)$ forms a rhombus. I tried to solve the problem by treating the complex numbers as vectors and working with dot products. But, that didn't seem to work. I also tried:

$|z+z^{2}| = |z(1+z)| = |z||1+z| = 2$

But clearly, that is wrong.

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  • $\begingroup$ @ThomasAndrews I believe so because the question just says "$z = \cos(\theta)+i\sin(\theta)$" So, $r = 1$ $\endgroup$ – NoLand'sMan Sep 19 at 14:47
  • $\begingroup$ If $|z|=1$ then $0\le|1+z|\le2; \;|1+z|=2$ holds only when $z=1$ $\endgroup$ – J. W. Tanner Sep 19 at 15:33
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$z=e^{i\theta}$, then $$|z+z^2|=|e^{i \theta}||1+\cos \theta+i \sin \theta|=\sqrt{(1+\cos \theta)^2+\sin ^2 \theta}=\sqrt{2+2\cos\theta}=2 \cos(\theta/2).$$

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Set $\theta=2y$

Use double angle formula $$1+z=1+\cos2y+sin2y=2\cos y(\cos y+i\sin y)$$

$|1+z|=|2\cos y||\cos y+i\sin y|=?$

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If $z=e^{i\theta}$ then $|z+z^2|=|z||1+z|=|e^{i\theta}||1+e^{i\theta}|=1|e^{i\theta/2}||e^{-i\theta/2}+e^{i\theta/2}|=2\mid\cos\frac{\theta}2\mid.$

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HINT:

Define $z=cis(\theta)$, this leads to de-Moivre's theorem: $cis^{a}(\theta)=cis(a\theta)$. Now, to any imaginary number it's norm is defined by: $\sqrt{\mathfrak{R}^{2}(z)+\mathfrak{I}^{2}(z)}$, where $\mathfrak{R}$ and $\mathfrak{I}$ are the real and imaginary parts, what are they in your case?

After that you'll need a few trig-identities: $$\cos{^{2}\theta}+\sin{^{2}\theta}=1$$ $$\cos (\theta_1-\theta_2)=\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2$$ $$\cos 2 \theta=2 \cos ^{2} \theta-1$$

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