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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
 #1
avatar+313 
+1

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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0

Sorry is wrong i dont know answer online

 
Guest Jan 12, 2019
edited by Guest  Jan 12, 2019
 #3
avatar+94995 
+2

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    \(\sqrt2\cdot i\)

.
 
 Jan 12, 2019

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