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I KNOW it can be solved by the trig formula, but I want to solve it by the square root definition, so please don't just post an alternative way to do it.

By the square root definition:

$$z = 5-12i$$ $$\sqrt{z} = w\implies w^2 = z$$

So if I suppose $w = a+bi$ we have:

$$w^2 = z \implies (a+bi)^2 = 5-12i\implies\\a^2-b^2+2abi = 5-12i\implies \\5 = a^2-b^2\\-12 = 2ab\implies$$

$$b^4-5b^2-36 = 0\implies b = \pm \sqrt{-4} = \pm 2i, b = \pm 3$$

Then I get $4$ solutions: For $b = \pm 2i$ I get $a = \pm 3i$ then $\sqrt{z}$ is $$\pm 2 + 3i$$ For $b = \pm 3$ I get $a = \pm 2$ and $\sqrt{z}$ is $$2\pm 3i$$

But two of these, when squared, aren't $z$. What am I doing wrong?

Guerlando OCs
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1 Answers1

6

An easier way is to solve: $$\begin{cases}a^2-b^2=5\ \ (1)\\ 2ab=-12\ \ (2)\\a^2+b^2=|5-12i|=13\ \ (3),\end{cases}$$

because you don't have to solve a equation of degree $4$.

Therefore, by $(1)$ $$a^2=b^2+5$$

By $(3)$ $$2b^2=8\implies b^2=4\implies b=\pm2$$

and by $(2)$ $$ab=-6\implies a=\mp3$$

Therefore $$\sqrt{5-12i}=\mp3\pm 2i$$

Surb
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