Kinsley Moon

2022-02-01

How do you divide $\frac{4+2i}{1-i}$?

lorugb

Expert

You must eliminate the complex number in the denominator by multiplying by its conjugate:
$\frac{4+2i}{1-i}=\frac{\left(4+2i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}$
$\frac{4+4i+2i+2{i}^{2}}{1-{i}^{2}}$
$\frac{4+6i-2}{1+1}$
$\frac{2+6i}{2}$
$1+3i$

ocretz56

Expert

Require the denominator to be real. To achieve this multiply the numerator and the denominator by the complex conjugate of the denominator.
If $\left(a+bi\right)$ is a complex number then $\left(a-bi\right)$ is the conjugate here the conjugate of
now $\frac{\left(4+2i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}$
distribute the brackets to obtain:
$\frac{4+6i+2{i}^{2}}{1-{i}^{2}}$
note that ${i}^{2}=\left({\sqrt{-1}}^{2}\right)=-1$
Therefore, $\frac{4+6i-2}{1+1}=\frac{2+6i}{2}=\frac{2}{2}+\frac{6i}{2}=1+3i$

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