Finding $\sqrt{a+ib}$ How do to find the square root of a complex number?
$z=a+ib$
$z=\sqrt{3+i4}=a+ib$
remove the square by squaring both side
$a^2+2iab+b^2=3+i4$
$a^2+b^2=3$
$2ab=4$
So I just solve the simultaneous equation. But my friend is telling me this way I am doing is wrong. Can anyone give a hand, thank in advance
 A: Be careful : $(a+ib)^{2}=a^{2}-b^{2}+2iab$ !
A: The method is correct, you identify the real and imaginary parts:
$$\begin{cases}x^2-y^2=a,\\2xy=b.\end{cases}$$
Then multiplying the first equation by $4x^2$,
$$4x^4-4ax^2-4x^2y^2=4x^4-4ax^2-b^2=(2x^2-a)^2-a^2-b^2=0,$$
giving
$$x^2=\frac{\pm\sqrt{a^2+b^2}+a}2.$$
As we need $a$ to be real, only the $+$ sign is useful and
$$x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}2}.$$
A: You can make it shorter using the moduli: you eventually have to solve the system
$$\begin{cases}a^2+b^2=5\\a^2-b^2=3 \\ab=2\\\end{cases}\iff \begin{cases}a^2 =4\\b^2=1\\ab>0\end{cases}$$
The last condition means $a$ and $b$ have the same sign,, so the square roots of $3+4i$ are
$$\color{red}{\pm(2+i)}.$$
A: If you really want the square root of an arbitrary complex number, then solving simultaneous equations is great.... well, it's what you've got. If you happen to know that you're dealing with a complex number that has a "nice" square root (i.e., rational real and imaginary parts), then you can use some tricks.
The norm of $a+bi$ is given by $a^2+b^2$, and is a multiplicative function. Thus, if you want the square root of, say $-\frac7{12}+\frac59i$, start by checking its norm: $\left(\frac7{12}\right)^2+\left(\frac59\right)^2=\frac{49}{144}+\frac{25}{81}=\frac{841}{1296}$. That's equal to $\left(\frac{29}{36}\right)^2$, so the square root, assuming there's a nice one, has norm $\frac{29}{36}$. Looking at that, you might guess $\frac56$ and $\frac26$ as options for real/imaginary parts, in some order. To decide which one, I'd look at where $-\frac7{12}+\frac59i$ lies in the plane - it's along the $\Bbb R^2$ vector $(-21,20)$. Splitting that angle in half mentally, we expect something in the part of the first quadrant nearer the positive imaginary axis, so we try $\frac13+\frac56i$, and that works.
Again, this is not a general approach for any complex number, but for something that arises in a context where you expect its square root to be reasonably nice, and where the numbers aren't too messy, this is great.
A: Much easier: If you want to find $\sqrt{a+ib}$ start by writing $$a+ib=re^{it}$$for some $r>0$ and $t\in\mathbb R$.
