How do you simplify $\frac{2 + 4 i}{2 i}$ $(2+4i)/(2i)$ ?

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Answer 1 · 2016-09-10T02:39:43

$2 - i$ $2-i$

First, rewrite the denominator in $a + b i$ $a+bi$ form

$\frac{2 + 4 i}{0 + 2 i}$ $(2+4i)/(0+2i)$

Secondly, multiply by the numerator and denominator by the complex conjugate of the denominator as follows:

$(\frac{2 + 4 i}{0 + 2 i}) (\frac{0 - 2 i}{0 - 2 i})$ $((2+4i)/(0+2i))((0-2i)/(0-2i))$

Expand numerator and denominator.

$\frac{0 - 4 i + 0 i - 8 i^{2}}{0 - 0 i + 0 i - 4 i^{2}}$ $(0-4i+0i-8i^2)/(0-0i+0i-4i^2)$

Recall that $i^{2} = - 1$ $i^2=-1$

$\frac{- 4 i + 8}{4} = \frac{- 4 i}{4} + \frac{8}{4} = - i + 2 = 2 - i$ $(-4i+8)/4=(-4i)/4+8/4=-i+2=2-i$

How do you simplify (2+4i)/(2i)2+4i2i(2+4i)/(2i)?