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Let z and w be complex numbers satisfying $$|z| = 5$$,  $$|w| = 2$$ and $$z\overline{w} = 6+8i$$. Find in the numbers $$|z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2$$

Aug 7, 2020
edited by littlemixfan  Aug 7, 2020

#1
+111172
0

I really do not know what I am doing here. (I have forgotten it)

However.

$$|z|=5\;\;\quad so\;\; \qquad z=5e^{\theta i}\\ |w|=2\;\;\quad so\;\; \qquad w=2e^{\alpha i}\\ |\bar w|=2\;\;\quad so\;\; \qquad \bar w=2e^{-\alpha i}\\~\\ zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\ |zw|^2=100\\ \left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\ z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\ \theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\ z\bar w=6+8i=10e^{(atan(1.\dot 3))i}$$

I think that is right as far as it goes but it does not answer all your questions.

LaTex:

zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\
|zw|^2=100\\
\left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\
z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\
\theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\
z\bar w=6+8i=10e^{(atan(1.\dot 3))i}

Aug 8, 2020
#2
+31103
+3

Here's a starter:

See if you can take it from here.

Aug 8, 2020
#3
+111172
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Thanks Alan,

Melody  Aug 8, 2020