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.


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.


  • 3
    $\begingroup$ 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 $\,|z+w|^2 = (z+w)(\bar z + \bar w) = |z|^2+|w|^2 + 2 \text{Re}(z \bar w)\,$. $\endgroup$
    – dxiv
    Commented May 11, 2022 at 3:15
  • 3
    $\begingroup$ 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. $\endgroup$
    – Salcio
    Commented May 11, 2022 at 3:18
  • $\begingroup$ @dxiv Thanks for the feedback. One follow-up question... sorry if it's a silly one. How does one get $z\bar{w} + w\bar{z} = 2\Re(z\bar{w})$? $\endgroup$
    – Serafina
    Commented May 11, 2022 at 3:23
  • $\begingroup$ @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. $\endgroup$
    – Serafina
    Commented May 11, 2022 at 3:25
  • 1
    $\begingroup$ @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 $\,\bar a = \overline{z \bar w}=\bar z w\,$. $\endgroup$
    – dxiv
    Commented May 11, 2022 at 3:31


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