# Complex numbers: adding two fractions and solving for z

## Homework Statement

$$\frac{1}{z}+\frac{1}{2-z}=1$$

## The Attempt at a Solution

Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:

$$\frac{1}{z}+\frac{1}{2-z}=1 \Rightarrow \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)}=1 \Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1 \Rightarrow \frac{-2}{2z-2z^2}=1 \Rightarrow -2z^2+2z+2=0 \Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2} \Rightarrow \frac{-2\pm\sqrt{-12}}{-4} \Rightarrow \frac{2\pm4i\sqrt{3}}{4} \Rightarrow \frac{1\pm2i\sqrt{3}}{2} \Rightarrow \frac{1}{2}\pm i\sqrt{3}$$

And this is far from correct.. I should have gotten ## 1 \pm i##.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

$$\frac{1}{z}+\frac{1}{2-z}=1$$

## The Attempt at a Solution

Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:$$\frac{1}{z}+\frac{1}{2-z}=1 \Rightarrow \frac{z}{z(2-z}+\frac{z-2}{z(2-z)}=1 \Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1 \Rightarrow \frac{-2}{2z-2z^2}=1 \Rightarrow -2z^2+2z+2=0 \Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2} \Rightarrow \frac{-2\pm\sqrt{-12}}{-4} \Rightarrow \frac{2\pm4i\sqrt{3}}{4} \Rightarrow \frac{1\pm2i\sqrt{3}}{2} \Rightarrow \frac{1}{2}\pm i\sqrt{3}$$
And this is far from correct.. I should have gotten ## 1 \pm i##.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!
I noticed a couple of errors.

##\displaystyle \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)} \ ## is not equivalent to ##\displaystyle \ \frac{z+z-2}{2z-z^2+2z-z^2} \ ##

The two terms have a common denominator. Don't add the denominators.

Also;
What is the square root of ##\ -12 \ ##?

lamefeed
Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of ##-12## I forgot to do ## \sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}##

SammyS
Staff Emeritus
Homework Helper
Gold Member
Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of ##-12## I forgot to do ## \sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}##
Yup. Of course, once you correct the first error, you won't have a square root of ##\ -12## .

Consider it solved!

$$\frac{z+(-z)+2}{2z-z^2}=1 \Rightarrow \frac{2}{2z-z^2} \Rightarrow 2=2z-z^2 \Rightarrow z^2-2z+2=0 \Rightarrow \frac{2\pm\sqrt{2^2-4*1*2}}{2} \Rightarrow \frac{2\pm\sqrt{-4}}{2} \Rightarrow \frac{2\pm2i}{2} \Rightarrow \\ 1 \pm i$$

I got such a good feeling when I solved it. Feels weird, i'm 100% super excited! Thanks SammyS and fresh_42(reply in the old thread).

You can optimise
$$\frac{1}{z} + \frac{1}{2-z} = 1 \iff \frac{1}{z} = 1 + \frac{1}{z-2} \iff \frac{1}{z} = \frac{z-1}{z-2} \iff z^2 -2z +2 =0.$$
and you get solutions ##1\pm i ##. If you have something of the form ##\frac{a}{f(x)} + \frac{b}{g(x)} = c ##, perhaps it's not such a hot idea to immediately mess with the common denominators and such.