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Let $z=-8+15i$ and $w=6-8i.$ Compute \[\dfrac{z\overline z}{w\overline w},\]where the bar represents the complex conjugate.

 Jun 24, 2019
 #1
avatar+101813 
+2

\(\dfrac{z\overline z}{w\overline w}\)

 

So we have

 

[-8 + 15 i ] [ -8 - 15i ]               64  - 225i^2            64 - 225(-1)           289

_________________  =       ___________  =    ___________  =    ______

[ 6 - 8i ] [ 6 + 8i ]                     36 - 64i^2              36  - 64(-1)              100

 

 

cool cool cool

 Jun 25, 2019
 #2
avatar
0

I am confused. What?

 Jun 25, 2019
 #3
avatar+8406 
+2

This video really helped me understand: https://www.youtube.com/watch?v=BZxZ_eEuJBM

 

 

If     \(z=-8+15i\)     then     \(\overline{z}=-8-15i\)

 

If     \(w=6-8i\)     then     \(\overline{w}=6+8i\)

 

 

And so...

 

 

\(\dfrac{z\overline{z}}{w\overline{w}}\ =\ \dfrac{(-8+15i)(-8-15i)}{(6-8i)(6+8i)}\ =\ \dfrac{(-8)^2-(15i)^2}{(6)^2-(8i)^2}\ =\ \dfrac{64+225}{36+64}\ =\ \dfrac{289}{100}\)

 

 

Just like CPhill found. laugh

 Jun 25, 2019
edited by hectictar  Jun 25, 2019
 #4
avatar+22527 
+3

Let

\(z=-8+15i \text{ and }w=6-8i.\)

 

Compute

\(\dfrac{z\overline z}{w\overline w}\),

where the bar represents the complex conjugate.

 

\(\begin{array}{|rcl|rcl|} \hline z\overline z &=& |z|^2 & w\overline w &=& |w|^2 \\ \hline z &=&-8+15i & w&=&6-8i \\ |z| &=& \sqrt{8^2+15^2} & |w| &=& \sqrt{6^2+8^2} \\ |z|^2 &=& 8^2+15^2 & |w|^2 &=& 6^2+8^2 \\ \hline \mathbf{\dfrac{z\overline z}{w\overline w}} &=& \dfrac{|z|^2}{|w|^2} \\ &=& \dfrac{8^2+15^2}{6^2+8^2 } \\ &=& \mathbf{ \dfrac{289}{100} } \\ \hline \end{array} \)

 

laugh

 Jun 26, 2019

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