Question #89196

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Answer 1 · 2016-11-24T00:57:08

$z_1= 2e^((3pi)/2i)$
$z_2 =8e^((4pi)/3i)$
$z_1z_2 = 16e^((5pi)/6i)$
$z_1/z_2=1/4e^(pi/6i)$

Using Euler's formula $e^(itheta) = cos(theta)+isin(theta)$ .

$z_1 = -2i$

$= 2(-i)$
$= 2(cos((3pi)/2)+isin((3pi)/2))$
$= 2e^((3pi)/2i)$

$z_2 = -4-4sqrt(3)i$

$= 8(-1/2-sqrt(3)/2i)$
$= 8(cos((4pi)/3)+isin((4pi)/3))$
$=8e^((4pi)/3i)$

Next, using that $e^xe^y = e^(xy)$ and $e^x/e^y = e^(x-y)$ , we have

$z_1z_2 = 2e^((3pi)/2i)*8e^((4pi)/3i) = 16e^((3/2+4/3)pii)=16e^((17pi)/6i) = 16e^((5pi)/6i)$

where the last step uses $e^(itheta) = e^(i(theta+2pin)), n in ZZ$

and

$z_1/z_2 = (2e^((3pi)/2i))/(8e^((4pi)/3i)) = 1/4e^((3/2-4/3)pii)=1/4e^(pi/6i)$