Question #89196

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Question #89196

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Answer 1 · 2016-11-24T00:57:08

$z_{1} = 2 e^{\frac{3 π}{2} i}$ $z_1= 2e^((3pi)/2i)$
$z_{2} = 8 e^{\frac{4 π}{3} i}$ $z_2 =8e^((4pi)/3i)$
$z_{1} z_{2} = 16 e^{\frac{5 π}{6} i}$ $z_1z_2 = 16e^((5pi)/6i)$
$\frac{z_{1}}{z_{2}} = \frac{1}{4} e^{\frac{π}{6} i}$ $z_1/z_2=1/4e^(pi/6i)$

Explanation:

Using Euler's formula $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta)+isin(theta)$ .

$z_{1} = - 2 i$ $z_1 = -2i$

$= 2 (- i)$ $= 2(-i)$
$= 2 (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $= 2(cos((3pi)/2)+isin((3pi)/2))$
$= 2 e^{\frac{3 π}{2} i}$ $= 2e^((3pi)/2i)$

$z_{2} = - 4 - 4 \sqrt{3} i$ $z_2 = -4-4sqrt(3)i$

$= 8 (- \frac{1}{2} - \frac{\sqrt{3}}{2} i)$ $= 8(-1/2-sqrt(3)/2i)$
$= 8 (cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3}))$ $= 8(cos((4pi)/3)+isin((4pi)/3))$
$= 8 e^{\frac{4 π}{3} i}$ $=8e^((4pi)/3i)$

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

$z_{1} z_{2} = 2 e^{\frac{3 π}{2} i} \cdot 8 e^{\frac{4 π}{3} i} = 16 e^{(\frac{3}{2} + \frac{4}{3}) π i} = 16 e^{\frac{17 π}{6} i} = 16 e^{\frac{5 π}{6} i}$ $z_1z_2 = 2e^((3pi)/2i)*8e^((4pi)/3i) = 16e^((3/2+4/3)pii)=16e^((17pi)/6i) = 16e^((5pi)/6i)$

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

and

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

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Question #89196

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