Complex numbers: Understanding solutions to tough problems

[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,
<br /> |1+z| &lt; \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.<br />
Does the "or" in there mean that both of those conditions must be met?

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

We obtain
<br /> (1+a^{2}-b^{2})^{2} + 4a^{2}b^{2}&lt;1 \text{ and } (1+a)^{2}+b^{2}&lt;\frac{1}{2},<br />
Did this come from just inserting z^{2}? I don't get this when inserting.
and consequently
<br /> (a^{2}+b^{2})^{2}+2(a^{2}-b^{2})&lt;0 \text{ and } 2(a^{2}+b^{2})+4a+1&lt;0.<br />
This must just be further reduction, but I need the previous step to understand this one.
Summing these inequalities implies
<br /> (a^{2}+b^{2})^{2}+(2a+1)^{2}&lt;0,<br />
which is a contradiction.
Problem 2:
Prove that
<br /> \sqrt{\frac{7}{2}} \leq |1+z| + |1-z+z^{2}| \leq 3\sqrt{\frac{7}{6}}<br />
for all complex numbers with |z|=1.

Solution:
Let t=|1+z|\in [0,2]. We have
<br /> t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.<br />
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
<br /> f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.<br />
No problemo.
We obtain
<br /> 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}}<br />
I understand this as well.

Any ideas?

 

[Mark44]

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,
<br /> |1+z| &lt; \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.<br />
Does the "or" in there mean that both of those conditions must be met?

No - "or" means that one or the other (or possibly both) must be true.

Solution:
Suppose by way of contradiction that
<br /> |1+z|&lt;\frac{1}{\sqrt{2}} \text{ and } |1+z^2|&lt;1.<br />

No. The opposite of the statement with "or" in it is
<br /> |1+z| \geq \frac{1}{\sqrt{2}} \text{ and } |z^2 + 1| &lt; 1.<br />
It's getting late, so I'm going to turn in, so maybe others will have additional comments on your work.

Setting z=a+bi, with a,b\in \mathbb{R} yields z^{2}=a^{2}-b^{2}+2abi. This step I understand.

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



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

Solution:
Let t=|1+z|\in [0,2]. We have
<br /> t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.<br />
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
<br /> f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.<br />
No problemo.
We obtain
<br /> 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}}<br />
I understand this as well.

Any ideas?

 

[HallsofIvy]

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,
<br /> |1+z| &lt; \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.<br />
Does the "or" in there mean that both of those conditions must be met?

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

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

We obtain
<br /> (1+a^{2}-b^{2})^{2} + 4a^{2}b^{2}&lt;1 \text{ and } (1+a)^{2}+b^{2}&lt;\frac{1}{2},<br />
Did this come from just inserting z^{2}? I don't get this when inserting.
and consequently
<br /> (a^{2}+b^{2})^{2}+2(a^{2}-b^{2})&lt;0 \text{ and } 2(a^{2}+b^{2})+4a+1&lt;0.<br />

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&lt; 1.

This must just be further reduction, but I need the previous step to understand this one.
Summing these inequalities implies
<br /> (a^{2}+b^{2})^{2}+(2a+1)^{2}&lt;0,<br />
which is a contradiction.



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

Solution:
Let t=|1+z|\in [0,2]. We have
<br /> t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.<br />
I know that Re(z)=\frac{z+\bar{z}}{2}, but isn't the equation missing z\cdot \bar{z}?

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.

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
<br /> f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.<br />
No problemo.
We obtain
<br /> 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}}<br />
I understand this as well.

Any ideas?

 

[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?