# Show that $|z+z^2|=2 \cos \frac{\theta}{2}$

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

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.

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

$$z=e^{i\theta}$$, then $$|z+z^2|=|e^{i \theta}||1+\cos \theta+i \sin \theta|=\sqrt{(1+\cos \theta)^2+\sin ^2 \theta}=\sqrt{2+2\cos\theta}=2 \cos(\theta/2).$$

Set $$\theta=2y$$

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

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

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

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: $$\cos{^{2}\theta}+\sin{^{2}\theta}=1$$ $$\cos (\theta_1-\theta_2)=\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2$$ $$\cos 2 \theta=2 \cos ^{2} \theta-1$$