Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I solve the following equation: $z/w$
$z= 5+5i$ and
$ w =2-i$

share|cite|improve this question
Can I ask why you have not accepted answers on any of the questions you have asked yet? I appreciate that you are new to the site, but it is polite if people to take time to answer you to reward them for their efforts. – Simon Hayward Oct 30 '12 at 16:49
Details here if needed – Simon Hayward Oct 30 '12 at 16:50
I tried to give plus reputation and click on the arrow - up As i really appreciated all the answers from u guys in here :). But it said I lacked reputation to do so. I simply though that was the only way to mark, how i appreciated. But thanks for elaborating it for me, I found the "function: accept answer". As I wasn't aware of the other function, until u just mentioned. – Alek Oliver Oct 30 '12 at 16:59
Ok cool. Yeah, you can always accept an answer and it boosts your reputation to do so :) – Simon Hayward Oct 30 '12 at 16:59
And now you can upvote comments and answers too... – Simon Hayward Oct 30 '12 at 17:01
up vote 4 down vote accepted

When dividing complex numbers the way to do it is to multiply by the conjugate on the top on bottom, so the bottom will become real.

$\frac{z}{w} = \frac{z\overline{w}}{w\overline{w}}$

In this case you have $$\frac{(5 + 5i)(2 +i)}{(2+i)(2-i)} = \frac{(5+5i)(2+i)}{5}$$

So now that the bottom is real I'm sure you can solve it!

share|cite|improve this answer
So i get $10+5i+10i+5i*i$=$15i+5$ Then we divide with 5 and get the result = $1+3i$ Thanks alot – Alek Oliver Oct 30 '12 at 17:25
@AlekOliver Yeah thats right, glad I could help. – Deven Ware Oct 30 '12 at 17:47

You first need to find $$w^{-1}=(2-i)^{-1}$$

This is $$\frac{\bar w }{|w|^2}$$

which is $$\frac{2+i}{5}$$

Then you can easily find $$zw^{-1}$$

ADD For any complex number $w=a+bi\neq 0$ we define its modulus as $$|w|=\sqrt {a^2+b^2}$$ and it's conjugate as $$\bar w =a-bi$$

Note that $$w\bar w =a^2+b^2=|w|^2$$

so we can conclude that for every $w\neq 0$, $$w^{-1}=\frac{\bar w }{|w|^2}$$

share|cite|improve this answer

Multiply the top and bottom by the complex conjugate of $w$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.