Let \(z\) and \(w\) be complex numbers satisfying \(|z| = 5, |w| = 2,\) and \(z\overline{w} = 6+8i.\) Then enter in the numbers \(|z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2 \), in the order listed above.
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\(\text{Let } z=a+bi\text{ and }w=c+di\\\text{Then }|z|=\sqrt{a^2+b^2}\text{, }|w|=\sqrt{c^2+d^2} \text{ and }z\bar w=(a+bi)(c-di)=ac+bd+(bc-ad)i\)
\(\text{Hence }a^2+b^2=5^2,ac+bd=6,c^2+d^2=2^2,bc-ad=8\)
(1)
\(|z+w|=|a+c+(b+d)i|=\sqrt{a^2+c^2+2ac+b^2+d^2+2bd}=\sqrt{a^2+b^2+c^2+d^2+2(ac+bd)}\\=\sqrt{5^2+2^2+2\times6}=\sqrt{41}\\\text{So }|z+w|^2=41\)
See if you can do the rest using a similar approach.
