Working complex fractions without conjugate method

[LF07LAN]

Homework Statement

Ive been given a question that requires an answer in polar form but the method I must use is the normal addition/subtraction of fractions. This throws me because I'm sure there is a simple method for the reciprocal of a complex number.

Q. 1/(13- 5i) - 1/(2-3i)

Homework Equations

where z = a + bi
i'm aware of how to change to polar for whole numbers and how to get where I want using the conjugate method but can't fathom the correct start point.

The Attempt at a Solution

I considered 1/13 - 1/5i using the rules of fractions but wasn't convinced.
The alternative was to convert straight to polar using 1 / 0degrees over the (13-5i in polar form)

 

[HallsofIvy]

\frac{1}{a+ bi}= \frac{1}{a+ bi}\frac{a- bi}{a- bi}= \frac{a- bi}{(a+ bi)(a- bi)}
= \frac{a- bi}{a^2-b^2i^2}= \frac{a- bi}{a^2+ b^2}

(Even for real numbers,
\frac{1}{a+ b}\ne \frac{1}{a}+ \frac{1}{b})

 

[LF07LAN]

Doesn't that just proove the Complex Conjugate method? I can only solve the answer without the use of this method.

The actual wording of the question is:
Evaluate the following expression using normal addition/subtraction of fractions and/or polar form (i.e DO NOT use the complex conjugate method) giving the result in polar form.

if I use (a-bi)/(a2 + b2) doesn't this just mean I've used the complex Conjugate Method? Or ar you saying I can do it this way:

where a^2 + b^2 = 13^2 + 5^2 = 194

13/194 + (5/194)i

0.067 + 0.026i
and the same for the second rectangular element of the question (1/(2-3i)

 

[Redbelly98]

Welcome to PF.

I think "normal addition/subtraction of fractions" means you find a common denominator for the two fractions, and then you can add the resulting numerators.