# Find real part of messy complex fraction?

I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?

$$\frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,$$

where $$R, X\in \mathbb R$$.

• Well, $\Re(300)=300$ – gammatester Nov 13 '18 at 10:39
• Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator? – saulspatz Nov 13 '18 at 10:43

## 2 Answers

I believe that R=resistance and X=impedance.

So:

$$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$$

$$\begin{cases} R277−X115 = 300 (R+277) \\ R115+X277 = 300 (115+X) \\ \end{cases}$$

$$\begin{cases} R277−X115 = 300R+300*277 \\ R115+X277 = 300*115+300X \\ \end{cases}$$

$$\begin{cases} R277-300R-X115 = 300*277 \\ R115+X277-300X = 300*115 \\ \end{cases}$$

$$\begin{cases} -23R-115X = 300*277 = 83100 \\ 115R+577X = 300*115 = 34500 \\ \end{cases}$$

$$\begin{cases} R=-1128613 \\ X= 225000 \\ \end{cases}$$

But R is negative! Can anybody verify ?

I'm assuming that you flipped right and left (since the real part of $$300$$ is clearly $$300$$). In general, to put a complex fraction in standard $$x+yi$$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:

$$\frac{a+bi}{c+di} ~~=~~ \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} ~~=~~ \frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$

Therefore, the real part of this fraction is

$$\frac{ac+bd}{c^2+d^2}.$$

In your case:

• $$a=R277-X115$$
• $$b=R115+X277$$
• $$c=R+277$$
• $$d=X+115$$

Plugging these into the formula above, we get

$$\frac{[R^2277-RX115+R277^2-X(277\cdot115)]+[RX115+X^2277+R115^2+X(277\cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$

Cancelling terms in the numerator gives

$$=\frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$

This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $$115^2=13225$$, and so on.

In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.

The resulting system is best solved by machine: here's the computation in WolframAlpha.