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I'm not sure how to figure this out (how to solve for $z$):

$$|z|=z+\bar{z}$$

What I did was,

Let $$z=a+bi $$ $$\sqrt{a^2+b^2}=(a+bi)+(a-bi)$$ $$\sqrt{a^2+b^2}=2a$$ $$a^2+b^2=4a^2$$

I'm not sure what to do from here to solve for $z$...

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Go on!

You get $$b^2=3a^2\implies b=\pm a\sqrt3$$ so the equation holds for all the complex numbers satisfying $$z=a\pm a\sqrt3i$$ due to conjugacy, with $a>0$ because $|z|$ must be non-negative.

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  • $\begingroup$ Could also have $-a\sqrt{3}$, no? $\endgroup$ – Randall May 29 '18 at 14:07
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    $\begingroup$ Note that $a>0$. $\endgroup$ – Szeto May 29 '18 at 14:14
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$$b^2+a^2=4a^2\to b^2=3a^2\to b=\pm a\sqrt3$$ Then filter that back in for $$z=a\pm i\cdot a\sqrt3 $$

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Another fun approach is the following:

Let $z=re^{i\theta}$. Then $\Re(z)=r\cos(\theta)$. Substituting this into the equation gives $$ r=2r\cos(\theta). $$ Therefore, either $r=0$ or $\cos(\theta)=\frac{1}{2}$. Hence $\theta=\frac{\pi}{3}$ or $\theta=-\frac{\pi}{3}$. Then, $$ z=re^{\frac{i\pi}{3}}\quad\text{or}\quad z=re^{-\frac{i\pi}{3}} $$ If you want to match more of the details, then use that $$ z=re^{\frac{i\pi}{3}}=r\cos\left(\frac{\pi}{3}\right)+ir\sin\left(\frac{\pi}{3}\right)=\frac{r}{2}+\frac{ir\sqrt{3}}{2}. $$

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Just some geometrical motivation:

enter image description here

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$$a^2+b^2=4a^2\implies b^2=3a^2 \implies b=\pm \sqrt 3 a $$

$$ z= a+bi = a\pm i\sqrt 3 a = a( 1\pm i\sqrt 3 ) $$

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