0
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ Oct 30, 2012 at 16:49
  • $\begingroup$ Details here if needed math.stackexchange.com/faq#howtoask $\endgroup$ Oct 30, 2012 at 16:50
  • $\begingroup$ 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. $\endgroup$ Oct 30, 2012 at 16:59
  • 1
    $\begingroup$ Ok cool. Yeah, you can always accept an answer and it boosts your reputation to do so :) $\endgroup$ Oct 30, 2012 at 16:59
  • $\begingroup$ And now you can upvote comments and answers too... $\endgroup$ Oct 30, 2012 at 17:01

3 Answers 3

4
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ So i get $10+5i+10i+5i*i$=$15i+5$ Then we divide with 5 and get the result = $1+3i$ Thanks alot $\endgroup$ Oct 30, 2012 at 17:25
  • $\begingroup$ @AlekOliver Yeah thats right, glad I could help. $\endgroup$
    – Deven Ware
    Oct 30, 2012 at 17:47
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .