How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?

SQUARE ROOT OF A BINOMIC COMPLEX NUMBER
The number $\sqrt{a+bi}$ is a complex (or complex) $x+yi$ such that: $a+bi=(x+yi)^{2}$
So: $a+bi=(x^{2}-y^{2})+2xyi\rightarrow \left.
x^{2}-y^{2}=a \atop
2xy=b
\right\}$ Solving this system, and taking into account, to solve the bi-square equation, that $\sqrt{a^{2}+b^{2}}\geq a$:
$$\left.
y=\dfrac{b}{2x} \atop
4x^{4}-4ax^{2}-b^{2}=0
\right\}\rightarrow \left.
x^{2}-y^{2}=a \atop
x^{2}=\dfrac{a+\sqrt{a^{2}+b^{2}}}{2}
\right\}\rightarrow \left.
x^{2}=\dfrac{a+\sqrt{a^{2}+b^{2}}}{2} \atop
y^{2}=\dfrac{-a+\sqrt{a^{2}+b^{2}}}{2}
\right\}$$
Now, the equation $2xy=b$ tells us that the product $xy$ has the same sign as $b$. Therefore, if $b>0$, $x$ and $y$ have the same signs, and if $b<0$, they have different signs.
$$b\geq 0\rightarrow \sqrt{a+bi}=\pm \left( \sqrt{\dfrac{a+\sqrt{a^{2}+b^{2}}}{2}}+i\sqrt{\dfrac{-a+\sqrt{a^{2}+b^{2}}}{2}}\right)$$
$$b<0\rightarrow \sqrt{a+bi}=\pm \left( \sqrt{\dfrac{a+\sqrt{a^{2}+b^{2}}}{2}}-i\sqrt{\dfrac{-a+\sqrt{a^{2}+b^{2}}}{2}}\right)$$
In practice, these formulas are not used, but the process is followed.
Also, it is highly recommended to pass it to polar.
