How to calculate $\sqrt{\frac{-3}{4} - i}$ 
Possible Duplicate:
How do I get the square root of a complex number? 

I know that the answer to $\sqrt{\dfrac{-3}{4} - i}$ is $\dfrac12 - i$. But how do I calculate it mathematically if I don't have access to a calculator? 

For $a \ne 0$, there are two numbers $z$ such that $z^2=a$.
We look at the given example, using only basic tools. We want to solve the equation
$$z^2=-\frac{3}{4}-i.$$
Because of a general discomfort with negative numbers, we look at the equivalent equation
$$z^2=-\frac{1}{4}(3+4i).$$
In order to deal with simpler numbers, we first consider  the equation
$$w^2=3+4i.$$
Let $w=a+ib$. Then $w^2=(a^2-b^2)+2ab i$.
So we want to solve the system of two equations
$$a^2-b^2=3, \qquad 2ab=4.$$
The solutions can be found by inspection. However, we continue, in order to show how to proceed when the numbers are less simple.  Rewrite the second equation as $b=2/a$. Substitute in the first equation. We get 
$$a^2-\left(\frac{2}{a}\right)^2=3,$$
which after some simplification becomes
$$a^4-3a^2-4=0.$$
This is a quadratic equation in $a^2$. By using the Quadratic Formula, we find that the roots are $a^2=4$ and $a^2=-1$. The second equation has no real solution, so $a=\pm 2$. We get the two solutions $a=2$, $b=1$ and $a=-2$, $b=-1$.  
Thus $w=2+i$ or $w=-(2+i)$. So find $z$, multiply these two values of $w$ by $\dfrac{i}{2}$.
Another way: Any complex number $z^2$ can be written as $r(\cos \theta+i\sin\theta)$ where $r$ is non-negative. Then 
$$z^2=r^2[(\cos^2\theta)+i(2\cos \theta\sin\theta)].$$
We can rewrite this as $r^2(\cos(2\theta)+i\sin(2\theta)$. We want $z^2=-\frac{3}{4}-i$. The norm of $-\frac{3}{4}-i$ is the square root of $(-3/4)^2+(-1)^2$, so $r^2=(9/16)+1=25/16$ and $r=5/4$.  We know that $\cos(2\theta)=(4/5)(-3/4)$ and $\sin(2\theta)=(4/5)(-1)$.
Simplify. We get $\cos(2\theta)=-\frac{3}{5}$ and $\sin(2\theta)=-\frac{4}{5}$. Now we could proceed by calculator, finding $2\theta$, then $\theta$, then $\cos\theta$ and $\sin\theta$.  Or else we can proceed algebraically, using the fact that $\cos^2\theta-\sin^2\theta=-\frac{3}{5}$ and $2\sin\theta\cos\theta=-\frac{4}{5}$. If we do this, the rest is much like the first solution. We have $\sin\theta=(-4/5)/(2\cos\theta)$. Substitute in $\cos^2\theta-\sin^2\theta=-\frac{4}{5}$. After simplifying, we obtain a quadratic equation in $\cos^2\theta$, and the rest is routine.  There will be two values of $\cos\theta+i\sin\theta$ that work, and they will be the negatives of each other, and the roots are $\pm\sqrt{r}(\cos\theta+i\sin\theta)$.
