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

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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
    
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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
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And now you can upvote comments and answers too... –  Simon Hayward Oct 30 '12 at 17:01

3 Answers 3

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!

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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 $w\neq 0$, $$w^{-1}=\frac{\bar w }{|w|^2}$$

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

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