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Solve this: $$z^2+i=|z|$$ I started playing the square then I high again at the square! $$(z^2+i)^2=|z|^2$$ but after that the calculations become many! there is a better way to solve it?

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  • $\begingroup$ Yes, much better. Write down the real and imaginary parts of both sides respectively. $\endgroup$
    – Lubin
    Oct 12, 2013 at 16:09
  • $\begingroup$ Have you tried a geometric approach? $\endgroup$
    – Jack M
    Oct 12, 2013 at 16:11

3 Answers 3

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Hint: $|z|^4 = |z^2|^2=\left||z|-i\right|^2 = |z|^2+1$.

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  • $\begingroup$ where did you get $|z|^4$? $\endgroup$
    – malloc
    Oct 12, 2013 at 16:08
  • $\begingroup$ $|z|^n = |z^n|$ for any $z \in \mathbb{C}$. $\endgroup$
    – njguliyev
    Oct 12, 2013 at 16:10
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HINT:

Let $z=r\cos(\theta+i\sin\theta)$

$\implies |z|=r$ and $z^2=\{r\cos(\theta+i\sin\theta)\}^2=r^2(\cos2\theta+i\sin2\theta)$

So we have, $r^2(\cos2\theta+i\sin2\theta)+i=r$

Equating the real & the imaginary parts

$r^2\cos2\theta=r\ \ \ \ (1)$ and $r^2\sin2\theta+1=0 \ \ \ \ (2)$

From $(1),$ either $r=0$ or $r\cos2\theta=1$

$r=0$ does not satisfy $(2)\implies r\cos2\theta=1\iff \cos2\theta=\frac1r$

From $(2)\sin2\theta=-\frac1{r^2}=-\cos^22\theta=\sin^22\theta-1$

Solve for $\theta$ from $\sin^22\theta-\sin2\theta-1=0$

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$$z^2 + i = |z|$$

$$(r \cos(\theta) + r i \sin(\theta))^2 + i = r$$

$$r^2\cos(2\theta) + r^2 i \sin(2\theta) + i = r$$

separate real and imaginary parts

$$r^2 \cos(2\theta) = r$$ $$r^2 \sin(2\theta) = -1$$

Squaring both sides introduces new solutions, so you'll have to double check the final result.

$$r^4 \cos(2\theta)^2 = r^2$$ $$r^4 \sin(2\theta)^2 = 1$$

Add the equations.

$$r^4 = r^2 + 1$$

Now you have a quadratic equation in r^2. It will have a few extra answers because of the squaring, so double check the result.

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