Equation with complex numbers and fractions I want to solve this, but I do not know how to continue after solving the square in the denominator. I tried multiplying by denominators, but I got awkward results.
I appreciate your help.

 A: What I'm thinking you want is to express $z_2$ as $A + Bi$, so you need to find $A$ and $B$.
To start on doing this, you'll need to make the denominators purely real.  This you do by multiplying out the denominators so they're in the form $C+Di.$  (The right term's denominator is already in this form: $C = \sqrt{3}$ and $D = -2$.)
Then, you multiply top and bottom of each term by the complex conjugate of the denominator in that term.  This gets rid of the $i$ in the denominator.
I'll do the right term:
$$\frac{\frac{1}{\sqrt{3}}}{ \sqrt{3}-2i} = \frac{1}{3 - 2i\sqrt{3}} = \frac{1}{3 - 2i\sqrt{3}}\left(\frac{3 + 2i\sqrt{3}}{3 + 2i\sqrt{3}}\right) = \frac{3 + 2i\sqrt{3}}{9 - 12i^2} = \frac{3 + 2i\sqrt{3}}{21}.$$
Do the same with the left term and you can simplify the expression.
A: You have two terms added. focus on one at a time, getting it into $u+iv$ format. To do this on your fractions, I'd start by performing the square in the denominator of the first one, so the denominator goes into $x+iy$ format, then multiply numerator and denominator of the first fraction by the conjugate $x-iy.$ This will give  real number denominator, and multiply out the numerator at this point, so the first term winds up in some kind of $u+iv$ form. A similar thing should be done to the second one, and finally adding at that point is immediate.
A: To simplify this I did the following. First expand out the square in the denominator.
$(1+i)^2=1+2i+i^2=2i\Rightarrow$
$z_2=\frac{-\frac{1}{\sqrt{3}}+2i}{5i+2i} + \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}-2i}$
Next multiply the second fractions numerator and denominator by the denominators conjugate.
$(\sqrt{3}-2i)(\sqrt{3}+2i)=3+4=7 \Rightarrow$
$z_2=\frac{-\frac{1}{\sqrt{3}}+2i}{7i} + \frac{\frac{\sqrt{3}+2i}{\sqrt{3}}}{7}=\frac{-\frac{1}{\sqrt{3}}+2i}{7i} + \frac{\sqrt{3}+2i}{7\sqrt{3}}$
Next we combine the fractions by multiplying them by the each others denominators.
$z_2=\frac{-7+14\sqrt{3}i+7\sqrt{3}i-14}{49\sqrt{3}i}=\frac{21\sqrt{3}i-21}{49\sqrt{3}i}=\frac{3\sqrt{3}i-3}{7\sqrt{3}i}$
Finally we simplify to get the answer.
$z_2=\frac{3}{7}-\frac{3}{7\sqrt{3}i}$
