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

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3 Answers 3


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!

  • $\begingroup$ 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

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


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


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