- #1

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Problem 1:

*Prove that for any complex number z,*

[tex]

|1+z| < \frac{1}{\sqrt{2}} \text{ or } |z^2 + 1| \geq 1.

[/tex]

**Does the "or" in there mean that both of those conditions must be met?**

Solution:

Suppose by way of contradiction that

[tex]

|1+z|<\frac{1}{\sqrt{2}} \text{ and } |1+z^2|<1.

[/tex]

Setting [itex]z=a+bi[/itex], with [itex]a,b\in \mathbb{R}[/itex] yields [itex]z^{2}=a^{2}-b^{2}+2abi[/itex].

**This step I understand.**

We obtain

[tex]

(1+a^{2}-b^{2})^{2} + 4a^{2}b^{2}<1 \text{ and } (1+a)^{2}+b^{2}<\frac{1}{2},

[/tex]

**Did this come from just inserting [itex]z^{2}[/itex]? I don't get this when inserting.**

and consequently

[tex]

(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0 \text{ and } 2(a^{2}+b^{2})+4a+1<0.

[/tex]

**This must just be further reduction, but I need the previous step to understand this one.**

Summing these inequalities implies

[tex]

(a^{2}+b^{2})^{2}+(2a+1)^{2}<0,

[/tex]

which is a contradiction.

Problem 2:

*Prove that*

[tex]

\sqrt{\frac{7}{2}} \leq |1+z| + |1-z+z^{2}| \leq 3\sqrt{\frac{7}{6}}

[/tex]

*for all complex numbers with*[itex]|z|=1[/itex].

Solution:

Let [itex]t=|1+z|\in [0,2].[/itex] We have

[tex]

t^{2}=(1+z)(1+\bar{z})=2+2Re(z), \text{ so } Re(z)=\frac{t^{2}-2}{2}.

[/tex]

**I know that [itex]Re(z)=\frac{z+\bar{z}}{2}[/itex], but isn't the equation missing [itex]z\cdot \bar{z}[/itex]?**

Then [itex]|1-z+z^{2}|=\sqrt{|7-2t^{2}|}[/itex].

**How did this happen? Inserting what I know about [itex]t[/itex] in [itex]|1-z+z^{2}|[/itex] doesn't get me anywhere.**

It suffices to find the extreme values of the function

[tex]

f:[0,2]\rightarrow \mathbb{R}, \, \, \, \, f(t)=t+\sqrt{|7-2t^{2}|}.

[/tex]

**No problemo.**

We obtain

[tex]

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}}

[/tex]

**I understand this as well.**

Any ideas?