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# Fast please help

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Let $z$ and $w$ be the complex numbers in the picture below:

If $|z| = 4$ and $|w| = 2$, then what is the imaginary part of $\dfrac{z}{w}$?​

Jan 12, 2019

### 3+0 Answers

#1
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ok just like the other one i did, the angle is 45 degrees. ignoring the diagram(again lol)

z=a+bi

w=c+di

a^2+b^2=16

c^2+d^2=4

if z=-4i then w is -2sqrt2-2isqrt2

divide w/z and take the reciprocal.

w/z = i(-2sqrt2-2isqrt2)/4 or (2sqrt2-2isqrt2)/4 which reduces to (sqrt2-isqrt2)/2.

to get z/w you have to take the reciprocal which is 2/(sqrt2-isqrt2). multiply by conjugate on top and bottom you get 2(sqrt2+isqrt2)/(2+2), which is

(sqrt2+isqrt2)/2 so the imaginary part is sqrt2/2.

HOPE THIS HELPED!!

Jan 12, 2019
#2
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Sorry is wrong i dont know answer online

Guest Jan 12, 2019
edited by Guest  Jan 12, 2019
#3
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If $|z| = 4$ and $|w| = 2$, then what is the imaginary part of $\dfrac{z}{w}$?​

$$w=2e^{i \theta}\\ z=4e^{ i(\theta+\frac{\pi}{4} )}\\~\\ \frac{z}{w}=\frac{4e^{ i(\theta+\frac{\pi}{4} )}}{2e^{i \theta}}\\ =2e^{i\theta+\frac{\pi i}{4}-i\theta}\\ =2e^{\frac{\pi i}{4}}\\ =2cos\frac{\pi}{4}+(2sin\frac{\pi}{4})i\\ =\frac{2}{\sqrt2}+\frac{2}{\sqrt2}i\\ =\sqrt2+\sqrt2i$$

So the imaginary part is    $$\sqrt2\cdot i$$

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Jan 12, 2019