Find z3 in the form of x+jy I would need to find 3z in the form of x+jy where:
$$\frac{1}{3z}=\frac{1}{(3-j4)}+\frac{1}{(3-j4)(5+j2)}$$
What I did was to expand the $$\frac{1}{(3-j4)(5+j2)}$$ which gives me $$\frac{1}{(23-14j)}$$
From here I am not very sure how to continue as I cannot seem to find a way to make both denominator the same.
Thanks for the help in advance!
 A: You can do standard arithmetic, finding a common denominator:
$$
\frac{1}{3z}=\frac{1}{(3-j4)}+\frac{1}{(3-j4)(5+j2)}=
\frac{(5+j2)+1}{(3-j4)(5+j2)}=
\frac{6+j2}{23-j14}
$$
Thus
$$
3z=\frac{23-j14}{6+j2}=\frac{23-j14}{6+j2}\frac{6-j2}{6-j2}=
\frac{110-j130}{36+4}=\frac{11}{4}-j\frac{13}{4}
$$
A: $$\frac{1}{23 - 14j} \cdot \frac{23 + 14j}{23 + 14j} = \frac{23 + 14j}{529 + 196} = \frac{23}{725} + \frac{14}{725}j$$
As this?
Clearly this is just an input for what you'll have to do. 
Use the rationalization method.
Namely: do the same with the first fraction $\frac{1}{3 - 4j}$ and you will get a similar thing. Then sum, simply.
A: Assuming $j^2=i^2=-1$:
$$\frac{1}{3z}=\frac{1}{(3-4i)}+\frac{1}{(3-4i)(5+2i)}\Longleftrightarrow$$
$$\frac{1}{3z}=\frac{1}{3-4i}+\frac{1}{23-14i}\Longleftrightarrow$$
$$\frac{1}{3z}=\frac{3}{25}+\frac{4i}{25}+\frac{23}{725}+\frac{14i}{725}\Longleftrightarrow$$
$$\frac{1}{3z}=\frac{22}{145}+\frac{26i}{145}\Longleftrightarrow$$
$$\frac{1}{\frac{1}{3z}}=\frac{1}{\frac{22}{145}+\frac{26i}{145}}\Longleftrightarrow$$
$$3z=\frac{11}{4}-\frac{13i}{4}$$
