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Let z and w be complex numbers satisfying |z| = 4 and |w| = 2. Find $$|z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2$$.

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

Dec 23, 2019

#1
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Tricky question.

For these, you need to remember the absolute value of a+bi = sqrt(a^2+b^2). So therefore we know that (this assumes that z = a+bi and 2 = c+di) sqrt(a^2+b^2) = 4. So a^2+b^2 is 16. Do the same with the other one, to get c^2+d^2 = 16. Now just put random numbers that work for that (they can be non-integers) and solve.

This is only for the first line, not the second. But you can do the same thing (but take into account the third equation w/ the conjugate).

Hope this helps! Dec 23, 2019
#2
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For the first problem, of course, |zw|^2 = |z|^2 |w|^2 = 64 and |z/w|^2 = |z|^2/|w|^2 = 4.

Then we can use the identities
|z + w|^2 = (z + wi)(z - wi) = |z|^2 + |z||w| + |w|^2 = 4^2 + 4*2 + 2^2 = 28
and
|z - w|^2 = (z - wi)(z + wi) = |z|^2 - |z||w| + |w|^2 = 4^2 - 4*2 + 2^2 = 12

For the second problem, |zw|^2 = |z|^2 |w|^2 = 100 and |z/w|^2 = |z|^2/|w|^2 = 25/4.

Then we can use the identities
|z + w|^2 = (z + wi)(z - wi) = |z|^2 + z bar(w) + bar(z) w + |z||w| + |w|^2 = 25 + (6 + 8i) + (6 - 8i) + 5*2 + 2^2 = 51
and
|z + w|^2 = (z + wi)(z - wi) = |z|^2 - z bar(w) - bar(z) w - |z||w| + |w|^2 = 25 - (6 + 8i) - (6 - 8i) - 5*2 + 2^2 = 7

Dec 23, 2019