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-1
645
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Let $z$ and $w$ be complex numbers satisfying $|z| = 5, |w| = 2,$ and $z\overline{w} = 6+8i.$ Then enter in the numbers \[|z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2 \]below, in the order listed above. If any of these cannot be uniquely determined from the information given, enter in a question mark.

 

I got 49 for the first one, 100 for the second one, 9 for the 3rd one, and 6.25 for the 4th one. But it was wrong, so I don't know how to do this question.

 Jun 28, 2020
edited by FearlessIsland3  Jun 28, 2020
 #1
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Just use the formula for |z|^2.

 Jun 28, 2020
 #2
avatar+118587 
+1

Clean up your question. Put your LaTex in a LaTex box  (found on the ribbon)

 Jun 29, 2020
 #3
avatar+9466 
+1

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

 

And then, we know that for any complex number z and w, \(z\overline{z} = |z|^2\) and \(|z||w| = |zw|\).

 

\(\quad|zw|^2 \\= |z|^2 |w|^2 \\= z\overline{z}w\overline{w} \\= (z\overline{w})(w\overline z) \\= (6 + 8i)(6 - 8i) \\= 10\)

 

\(\quad|z + w|^2 \\= (z + w)\left(\overline{z + w}\right) \\= (z + w)\left(\overline z + \overline w\right)\\ = z\overline z + w\overline w + z\overline w + w\overline z \\= |z|^2 + |w|^2 + z\overline w + \overline zw \\= 5^2 + 2^2 + (6 + 8i) + (6- 8i) \\= 41\)

 

I will leave the remaining ones to you.

 Jun 29, 2020
edited by MaxWong  Jun 29, 2020
edited by MaxWong  Jun 29, 2020

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