How can I solve the following equation: $z/w$
When
$z= 5+5i$ and
$ w =2-i$
Solving a division with imaginary numbers
0
$\begingroup$
complex-numbers
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0Thanks for the small introduction to how the system works :) – 2012-10-30
3 Answers
4
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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0@AlekOliver Yeah thats right, glad I could help. – 2012-10-30
3
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 $w\neq 0$, $w^{-1}=\frac{\bar w }{|w|^2}$
1
Multiply the top and bottom by the complex conjugate of $w$.