<s>[INST] How to find root of the complex number? When finding nth root of the complex number $z=\cos{x}+i\sin{x}$, we can use known De Moivre's formula:
$z^\frac1n=[r(\cos{x}+i\sin{x})]^\frac1n=r^\frac1n\left[\cos{\left(\frac{x+2k\pi}n\right)}+i\sin{\left(\frac{x+2k\pi}n\right)}\right]$
My task is to find $\sqrt{3+2i}$ without using De Moivre's formula, and my question is how to do that?
Thank you for your answers.
 [/INST] Note that if the root is $a+bi$ then $a^2-b^2=3$ and $2ab=2$
Then $a^2b^2=1$ so that $(a^2-b^2)^2+4a^2b^2=(a^2+b^2)^2=13$ and $a^2+b^2=\sqrt {13}$
Then it is easy to solve for $a^2$ and $b^2$, bearing in mind that $a$ and $b$ are real. Then you get $a$ and $b$, and fix the signs of the square roots to make the product come out with the right sign.
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