# Homework Help: Complex numbers: Understanding solutions to tough problems

1. Oct 17, 2011

### Hioj

Following are problems from the book "Complex Numbers from A to ...Z" by Titu Andreescu and Dorin Andrica. It's a wonderful book, I'm still adapting to the higher-than-usual level though. My questions/comments are written in bold throughout the problems and solutions.

Problem 1:
Prove that for any complex number z,
$$|1+z| < \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.$$
Does the "or" in there mean that both of those conditions must be met?

Solution:
Suppose by way of contradiction that
$$|1+z|<\frac{1}{\sqrt{2}} \text{ and } |1+z^2|<1.$$
Setting $z=a+bi$, with $a,b\in \mathbb{R}$ yields $z^{2}=a^{2}-b^{2}+2abi$. This step I understand.

We obtain
$$(1+a^{2}-b^{2})^{2} + 4a^{2}b^{2}<1 \text{ and } (1+a)^{2}+b^{2}<\frac{1}{2},$$
Did this come from just inserting $z^{2}$? I don't get this when inserting.
and consequently
$$(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0 \text{ and } 2(a^{2}+b^{2})+4a+1<0.$$
This must just be further reduction, but I need the previous step to understand this one.
Summing these inequalities implies
$$(a^{2}+b^{2})^{2}+(2a+1)^{2}<0,$$

Problem 2:
Prove that
$$\sqrt{\frac{7}{2}} \leq |1+z| + |1-z+z^{2}| \leq 3\sqrt{\frac{7}{6}}$$
for all complex numbers with $|z|=1$.

Solution:
Let $t=|1+z|\in [0,2].$ We have
$$t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.$$
I know that $Re(z)=\frac{z+\bar{z}}{2}$, but isn't the equation missing $z\cdot \bar{z}$?
Then $|1-z+z^{2}|=\sqrt{|7-2t^{2}|}$.How did this happen? Inserting what I know about $t$ in $|1-z+z^{2}|$ doesn't get me anywhere.
It suffices to find the extreme values of the function
$$f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.$$
No problemo.
We obtain
$$f\left(\sqrt{\frac{7}{2}}\right)=\sqrt{\frac{7}{2}} \leq t+\sqrt{|7-2t^{2}|} \leq f\left(\sqrt{\frac{7}{6}}\right)=3\sqrt{\frac{7}{6}}$$
I understand this as well.

Any ideas?

2. Oct 18, 2011

### Staff: Mentor

No - "or" means that one or the other (or possibly both) must be true.
No. The opposite of the statement with "or" in it is
$$|1+z| \geq \frac{1}{\sqrt{2}} \text{ and } |z^2 + 1| < 1.$$
It's getting late, so I'm going to turn in, so maybe others will have additional comments on your work.

3. Oct 18, 2011

### HallsofIvy

To this point, what Mark44 said: a contradiction would be
$$|1+ z^2|< 1\text{ and } |1+ z|\ge \frac{1}{\sqrt{2}}$$

No, it's not "just inserting $z^2$", it is using the definition of "absolute" value of a complex number: $|z|= \sqrt{z\bar{z}}$ so with z= a+ bi, $|a+ bi|= \sqrt{a^2+ b^2}$.

With z= a+ bi, 1+ z= (a+ 1)+ bi so the absolute value is $\sqrt{(a+1)^2+ b^2}\ge \frac{1}{\sqrt{2}}$. Squaring both side of the inequality $|1+ z|\ge \frac{1}{\sqrt{2}}$ (which is legal because both sides are positive) we have
$$(a+1)^2+ b^2\ge \frac{1}{2}$$

With z= a+ bi, $z^2= a^2-b^2+ 2abi$, $1+ z^2= a^2- b^2+ 1+ 2abi$ and $|1+ z^2|^2= (1+ a^2- b^2)^2+ 4a^2b^3< 1$.

The hypothesis is that |z|= 1 so that $z\cdot\bar{z}= 1$. In particular, with $t^2= |1+ z|^2$, again, taking z= a+ bi, $t^2= (a+1)^2+ b^2= a^2+ 2a+ 1+ b^2= 2a+ 2= 2Re(z)+ 2$ because $|z|^2= a^2+ b^2= 1$.

Last edited by a moderator: Oct 18, 2011
4. Oct 18, 2011

### Hioj

Ah! That makes so much more sense now. Looking at the problem with z=a+bi made much more sense. What happened when you went from $1+z^{2}$ to $|1+z^{2}|^{2}$? How did $a^{2}-b^{2}+1+2abi$ become $(1+a^{2}-b^{2})^{2}+4a^{2}b^{3}$?

What about the last part when the function showed up?