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Determine all possible values of z ∈ C that satisfy the equation $2z = (z^*)^2$. For real values I calculated 0 and 2 with simple algebra. But I'm not sure how to do this for complex numbers.
$2z = (z^*)^2$
$2(x+yi) = (x-yi)^2$
$0=x^2-2x-2ixy-2yi-y^2$
But I'm not sure if I'm supposed to solve it this way. (or how to solve it like this)

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2 Answers 2

up vote 2 down vote accepted

If $a+ib=c+id$ where $a,b,c,d$ are real

$a-c=i(d-b)$

Squaring we get $(a-c)^2=-(d-b)^2\implies (a-c)^2+(d-b)^2=0$

Now, sum of the squares of two real numbers is $0$

So, each must be $0$ as square of a real number $\ge 0$

$\implies a=c,d=b$

So, we can equate the real & the imaginary parts,

$$x^2-2x-y^2=0, 2xy+2y=0\implies 2y(x+1)=0$$

Either $y=0\implies x^2-2x=0\implies x=0,2\implies z=0+0\cdot i=0,2+0\cdot i=2$

or $x+1=0\implies x=-1\implies y^2=(-1)^2-2(-1)=3\implies y=\pm \sqrt 3$ $\implies z=-1\pm i\sqrt 3$

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Let $z = r e^{i \theta}$. Hence, we have $$2re^{i \theta}=r^2 e^{-2i \theta}$$ This gives us $$r^2 = 2r \, \text{ and }\, e^{3i\theta} = 1 \implies 3 \theta = 2n \pi \implies \theta \in \left\{\dfrac{2\pi}3, \dfrac{4 \pi}3, 2 \pi \right\}$$ For non-zero $z$, we get $r=2$. Hence, the possible values are $$z = 2 \cdot e^{2\pi i/3}, 2 \cdot e^{4\pi i/3}, 2 = -1 \pm i \sqrt3, 2$$

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