# Let $z, w$ be complex numbers lying in the first quadrant. Prove that $|z|^2 + |w|^2 \leq |z + w|^2 \leq 2(|z|^2 + |w|^2)$.

I'm starting to work my way through old complex analysis qualifying exams as a way to study for my upcoming exams in August. I'd like to know whether or not my proposed answer is sufficient and also whether or not there is a better way to do this. This is my first post on Math StackExchange, so I'm certainly open to learning better question formatting if needed as I will be posting plenty more over the summer as I study, both for feedback and help when I'm stuck.

Problem:

Let $$z, w$$ be complex numbers lying in the first quadrant. Prove that $$|z|^2 + |w|^2 \leq |z + w|^2 \leq 2(|z|^2 + |w|^2)$$.

Possible Solution:

Let $$z=a+bi$$ and $$w=c+di$$.

Then $$|z|^2 = a^2+b^2$$, $$|w|^2 = c^2+d^2$$, and $$|z+w|^2 = |(a+c) + (b+d)i|^2 = (a+c)^2 + (b+d)^2$$.

By substituting these into the given equality, we can rewrite it as $$a^2+b^2+c^2+d^2 \leq (a+c)^2 + (b+d)^2 \leq 2(a^2+b^2+c^2+d^2).$$

For the first inequality, we have

\begin{align*} a^2+b^2+c^2+d^2 &\leq a^2 + 2ac + c^2 + b^2 + 2bd + d^2 \\ 0 &\leq ac + bd \end{align*}

It is stated that both $$z$$ and $$w$$ lie in the first quadrant, thus $$a\geq 0, \;b\geq 0, \;c\geq 0,\; d\geq 0$$ and it's clear that $$ac + bd \geq 0$$ holds.

For the second inequality, we have \begin{align*} a^2 + 2ac + c^2 + b^2 + 2bd + d^2 &\leq 2(a^2+b^2+c^2+d^2) \\ 2ac + 2bd &\leq a^2 + b^2 + c^2 + d^2 \\ 2ac - a^2 - c^2 &\leq b^2 + d^2 - 2bd \\ -(a-c)^2 &\leq (b-d)^2 \end{align*}

Since both $$(a-c)^2$$ and $$(b-d)^2$$ must be positive, it is clear that $$-(a-c)^2 \leq (b-d)^2$$ holds as well.

$$\square$$

• This is a well written question (+1), and both proofs look correct. Just as a matter of style, I would have moved everything to one side for the second part i.e. $\,0 \le (a-c)^2+(b-d)^2\,$. Another possible way to approach it could be to start with $\,|z+w|^2 = (z+w)(\bar z + \bar w) = |z|^2+|w|^2 + 2 \text{Re}(z \bar w)\,$.
– dxiv
Commented May 11, 2022 at 3:15
• Also, note that for any complex numbers z and w one has $|z+w|^2 + |z-w|^2 = 2(|z|^2 + |w|^2)$. This is so called parallelogram law known from geometry. In fact, it is true for any scalar product norm. Commented May 11, 2022 at 3:18
• @dxiv Thanks for the feedback. One follow-up question... sorry if it's a silly one. How does one get $z\bar{w} + w\bar{z} = 2\Re(z\bar{w})$? Commented May 11, 2022 at 3:23
• @Salcio Thanks for pointing that out. I struggle to remember helpful things like this, but it would make quick work of the second part of the proof, to be sure. Commented May 11, 2022 at 3:25
• @Serafina $\;a + \bar a = 2 \text{Re}(a)\,$, $\,a - \bar a = 2i \text{Im}(a)\,$. Here it's the former with $\,a = z \bar w\,$ and $\,\bar a = \overline{z \bar w}=\bar z w\,$.
– dxiv
Commented May 11, 2022 at 3:31