Square roots of complex numbers I know that the square root of a number x, expressed as $\displaystyle\sqrt{x}$, is the number y such that $y^2$ equals x. But is there any simple way to calculate this with complex numbers? How?
 A: Sure. By the fundamental theorem of algebra, the complex numbers $\mathbb{C}$ are an algebraically closed field. Informally, what this means is that any polynomial equation with coefficients in $\mathbb{C}$ has roots in $\mathbb{C}$. Say, then, that you have a complex number $a+bi$ that you want to find the square root of. What you really want to do is find the solutions to the equation $x^{2} - (a+bi) = 0$ - you can see, for example, that $i$ is the solution to the equation $x^{2} - (-1) = 0$. 
The fundamental theorem of algebra tells us that our roots will be in $\mathbb{C}$, i.e., will be expressible as $c+di$ for some $c, d \in \mathbb{R}$. 
Plugging in, then, and equating real and imaginary parts, we see: 
$$(c+di)^{2} = a+bi$$
$$\implies c^{2} - d^{2} = a;~~ 2cd = b$$ 
For some fixed $a, b \in \mathbb{R}$, we can solve the equation thusly. 
Assuming $c \neq 0$, which certainly must be true in the case that $b \neq 0$ (in which case the answer is obvious), we note that:
$$d = \frac{b}{2c}$$
$$\implies c^{2} - (\frac{b}{2c})^{2} = a;$$ With some algebra, we see:
$$\implies c^{4} - 4ac^{2} - b^{2} = 0$$
Let $k = c^{2}$. Then this easily transforms to a quadratic which we can solve:
$$k^{2} - 4ak -b^{2} = 0;$$ Using the quadratic formula, then, we get:
$$k = \frac{4a \pm \sqrt{16a^{2} + 4b^{2}}}{2} = 2a \pm \sqrt{4a^{2} + b^{2}}$$
Note: the discriminant must be nonnegative, so $k$ is strictly real. This means that we don't have to worry about $c$ being the square root of a complex number, which would defeat the purpose of this method. 
Then, $$c = \pm\sqrt{k} = \pm \sqrt{2a \pm \sqrt{4a^{2} + b^{2}}}$$
and $$d = \frac{b}{2c}$$
A: If you represent the complex number $z$ in polar form, i.e.
$$z=re^{i\theta} = r(\cos \theta+i\sin\theta)$$
where $r>0, \theta \in [0, 2\pi)$.
Then the square roots of $z$ are
$$\sqrt z = \pm \sqrt re^{i\theta/2}$$
In general the $k$ $k$th roots of $z$ are $\sqrt[k]r\exp\left(i\times\left(\frac{2\pi j}{k}+\frac\theta k\right)\right)$ for $j=0,1,2,...,k-1.$
A: Write $(x+iy)^2=a+ib$, thus $x^2-y^2=a$ and $2xy=b$.
Now, $-x^2y^2=-b^2/4$, so $x^2$ and $-y^2$ are the solutions of $t^2-at-b^2/4=0$.
Then $\Delta=a^2+b^2$, and $t=\frac{a \pm \sqrt{a^2+b^2}}{2}$.
Obviously, since $x^2 \geq 0$ and $-y^2 \leq 0$, we have
$$x^2 = \frac{\sqrt{a^2+b^2}+a}{2}$$
$$y^2 = \frac{\sqrt{a^2+b^2}-a}{2}$$
Then, using the fact that $xy$ has the same sign than $b$, we have the two square roots of $a+ib$,
$$x+iy = \pm \left( \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} + i \ \mathrm{sign}(b) \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}\right)$$
with $\mathrm{sign}(x)=1$ if $x\ge0$ and $-1$ otherwise.
A: If you write the number $a+bi$ as $re^{i\phi}$ you can deduce: $\sqrt{a+bi}=\sqrt{re^{i\phi}}=e^{\pm \frac{i\phi}{2}}\sqrt{r}$
$r=\sqrt{a^2+b^2}$, $\phi=\arctan{\frac{b}{a}}$
