Knowing $|z|$ and $|w|$ (and possibly $z\overline{w}$), what are $|z+w|^2$, $|zw|^2$, $|z-w|^2$, $|z/w|^2$?

Question

Question 1: Let $z$ and $w$ be complex numbers satisfying $|z| = 4$ and $|w| = 2$. 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 plugged in $4$ and $2$ to each equation and got $36$, $64$, $4$, $4$, for question one. It was wrong.

Question 2: 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 don't know how to even start on question 2 because of the $z\overline{w}$.

Answer 1

For question one, use the definition of magnitude to help you:

$$|z|^2=\bar{z}z$$

Notice:

$$|z+w|^2=(z+w)(\bar{z}+\bar{w})\\=|z|^2+|w|^2+\bar{z}w+\bar{w}z$$

which, obviously, we don't have enough information to compute. However

$$\left|zw\right|^2=|z|^2|w|^2$$

which indeed we can determine. I will leave the other two problems for you to workout. This expansion will also be useful in solving question 2, where $z\bar{w}$ appears. As one final hint, notice that $\bar{z}w$ is the complex conjugate of $\bar{w}z$.

Answer 2

The mistake was plugging $|z|,|w|$ instead of $z,w$. For example, if $z=4,w=-2$, then $4=|z+w|^2 \neq ||z|+|w||^2=36$. Note that if $z=4,w=2$, we do get $36$. We conclude that the expression is not uniquely determined from the information given.

A way to deal with this expression would be: $|z+w|^2= (z+w) \overline{(z+w)} = |z|^2+|w|^2+z\overline{w}+\overline{z}w= 20+2|z||w| \cos(\theta)$, where $\theta$ is the angle between $z,w$. The expression does not depend only on the magnitudes of $z,w$- their angles make a difference.

The expression $|zw|^2= |z|^2|w|^2=64$, however, does depend only on the magnitudes. You can try the rest similarly, with the idea of where $z\overline{w} = \overline{\overline{z}w}$ comes from.

Answer 3

First, $|z| = 4$ doesn't even mean $z = 4$ in the real numbers. $z$ could be either $4$ or $-4$. In the complex numbers, $|z| = 4$ means that $z$ could be any point on the complex plane on the circle centered at $0+0\mathrm{i}$ having radius $4$. Similarly, $|w| = 2$ means $w$ could be any point on the complex plan on the circle centered at $0+0\mathrm{i}$ having radius $2$.

That is $z$ is a point on the blue circle and $w$ is a point on the orange circle.

Let's pick a point on the blue circle (pretty much at random) for $z$ and then let $w$ swing through all the values it is permitted to see all the points that could be $z+w$.

(Recall : this is one choice of $z$ and all the choices of $w$ shown simultaneously.)

If you draw the ray from the center of the blue circle out through the point $z$ we picked, it meets the orange circle of values of $z+w$ at two places. When $z$ and $w$ have opposite (antiparallel) complex angles, the point $z+w$ is the point closest to the origin where the ray and the orange circle intersect -- it has distance $4-2=2$ from the origin. When $z$ and $w$ have the same (parallel) complex angles, we get the other intersection, at distance $4+2=6$ from the origin.

As we let $z$ range over the circle, at each choice of $z$, we get a rotated version of the figure above, and so $$ 2 = 4-2 \leq |z+w| \leq 4+2 = 6 \text{.} $$ Just knowing the magnitudes of $z$ and $w$ does not tell you the magnitude of their sum.

Nor does knowing the magnitudes of $z$ and $w$ tell you the magnitude of their difference. The diagram looks the same, but we subtract $w$ instead of add.

There is an inequality, the triangle inequality that says $$ |z+w| \leq |z| + |w| \text{.} $$ Notice that this is an inequality -- the triangle inequality does not tell you the precise magnitude of the sum, only places an upper bound. The upper bound is attained when $z$ and $w$ have the same complex angles (are parallel). There is another inequality, sometimes called the reverse triangle inequality (that can be derived from the triangle inequality, so it is not really something new) that says $$ |z+w| \geq \left| \rule[0pt]{0pt}{10pt} |z| - |w| \right| \text{.} $$ This lower bound is attained when $z$ and $w$ have opposite (antiparallel) complex angles. If you check, you will see that the triangle inequality gives the same upper bound we got by looking at circles and the reverse triangle inequality gives the same lower bound.

However, multiplication and division are much better behaved for magnitudes. In particular, $$ |z||w| = |zw| \text{ and } \\ |z|/|w| = |z/w| $$
(assuming $w \neq 0$ for the division because division by zero is still not allowed).

So, $$ |zw|^2 = (|z||w|)^2 = |z|^2 |w|^2 $$ and you know both of $|z|$ and $|w|$. Similarly, $$ \left| \frac{z}{w} \right|^2 = \left( \frac{|z|}{|w|} \right)^2 = \frac{|z|^2}{|w|^2} $$ and, again, you know both of $|z|$ and $|w|$.

Moving on to problem 2...

If we express $z$ and $w$ in real and imaginary components, \begin{align*} z &= x + \mathrm{i}\, y \\ w &= u + \mathrm{i}\, v \text{,} \end{align*} where $x$, $y$, $u$, and $v$ are all real, then we can solve the equations $|z| = 5$, $|w| = 2$, and $z \overline{w} = 6+8\,\mathrm{i}$ for those components. Repeating the above, letting $z$ range around the circle of radius $5$ and for each choice of $z$ seeing which choice of $w$ satisfies $z \overline{w} = 6+8\,\mathrm{i}$, we obtain this plot.

Here we discover that the additional constraint on $z \overline{w}$ picks out an angle between the ray from the origin to $z$ and the ray from $z$ to $z+w$. All these $z+w$ lie on a circle, so $|z+w|$ is a fixed number so this part of question $2$ has a definite answer. Perhaps the easiest way to go is to set $z = 5+0\,\mathrm{i}$, find the components of $w$ for that point and compute $|(5+0\,\mathrm{i}) + (u+ \mathrm{i}\,v)|$ for that particular $w$.

The same general method will get you to $|z-w|$.

The same ideas as in problem $1$ get you $|zw|^2$ and $|z/w|^2$.