More problems from the lecturer's book:
#6D Show, for all values of $x$ and $a$, that$$3 x^{4}+(8 a-4) x^{3}+\left(6 a^{2}-12 a+6\right) x^{2}-\left(12 a^{2}+12\right) x+\left(7 a^{2}+6 a+9\right)>0$$What is the smallest value taken by the above expression as $x$ and $a$ vary?
Solution.
Let $f_{a}(x)=3 x^{4}+(8 a-4) x^{3}+\left(6 a^{2}-12 a+6\right) x^{2}-\left(12 a^{2}+12\right) x+\left(7 a^{2}+6 a+9\right)>0$.
Differentiating with respect to $x$ we get\begin{align*} f_{a}^{\prime}(x) &=12 x^{3}+(24 a-12) x^{2}+\left(12 a^{2}-24 a+12\right) x-\left(12 a^{2}+12\right) \\ &=12\left(x^{3}+(2 a-1) x^{2}+\left(a^{2}-2 a+1\right) x-\left(a^{2}+1\right)\right) \\ &=12(x-1)\left(x^{2}+2 a x+a^{2}+1\right) \\ &=12(x-1)\left\{(x+a)^{2}+1\right\} \end{align*}As $f'_a(x)$ goes from negative to positive at $x = 1$, then $f_a(x)$ has a minimum at $x = 1$ and this minimum is\begin{align*} f_{a}(1) &=3+(8 a-4)+\left(6 a^{2}-12 a+6\right)-\left(12 a^{2}+12\right)+\left(7 a^{2}+6 a+9\right) \\ &=a^{2}+2 a+2 \\ &=(a+1)^{2}+1>0 \end{align*}which answers the question. The smallest $f_a(x)$ can be is 1 when $x = 1$ and $a = −1$.
#7d How many real roots (including multiplicities) have the following polynomial equations?$$\left(x^{2}+1\right)^{10} =\left(2 x-x^{2}-2\right)^{7} ; \quad\left(x^{2}+2 x+3\right)^{8}=\left(2+2 x-x^{2}\right)^{5};\quad \left(x^{2}+1\right)^{5} =\left(1-2 x^{4}\right)^{7}$$Solution.
(i) LHS>0>RHS for all $x$ and so there are no real roots.
(ii) LHS≥256>243≥RHS for all $x$ and so there are no real roots.
(iii) LHS≥1≥RHS with equality only at $x = 0$ which is a double root.
#29B Show, for any complex numbers $z,w$, that $|1−\overline{z}w|^2−|z−w|^2 = (1−|z|^2)(1−|w|^2)$. Deduce that$$\left|\frac{z-w}{1-\bar{z} w}\right|=1\quad\text{if $|w|=1$ and $z\ne w$}$$Proof.\begin{align*}|1-\bar{z} w|^{2}-|z-w|^{2} &=(1-\bar{z} w)(1-z \bar{w})-(z-w)(\bar{z}-\bar{w}) \\ &=1-z \bar{z}-w \bar{w}+z \bar{z} w \bar{w} \\ &=(1-z \bar{z})(1-w \bar{w}) \\ &=\left(1-|z|^{2}\right)\left(1-|w|^{2}\right) \end{align*}#30C Show, for any distinct complex numbers $z,w$, that$$\operatorname{Re}\left(\frac{z+w}{z-w}\right)=\frac{|z|^{2}-|w|^{2}}{|z-w|^{2}}$$For given $w\ne0$, describe the locus of points $z$ that satisfy $\operatorname{Re}((z+w) /(z-w))=1$.
Proof.\begin{align*} \operatorname{Re}\left(\frac{z+w}{z-w}\right) &=\frac{1}{2}\left(\frac{z+w}{z-w}+\frac{\bar{z}+\bar{w}}{\bar{z}-\bar{w}}\right)\\ &=\frac{1}{2}\left(\frac{(z+w)(\bar{z}-\bar{w})+(z-w)(\bar{z}+\bar{w})}{(z-w)(\bar{z}-\bar{w})}\right) \\ &=\frac{1}{2}\left(\frac{2 z \bar{z}-2 w \bar{w}}{(z-w)(\bar{z}-\bar{w})}\right) \\ &=\frac{|z|^{2}-|w|^{2}}{|z-w|^{2}} \end{align*}So$$\operatorname{Re}\left(\frac{z+w}{z-w}\right)=1 \Longleftrightarrow \frac{|z|^{2}-|w|^{2}}{|z-w|^{2}}=1 \Longleftrightarrow|z|^{2}=|w|^{2}+|z-w|^{2}$$This means Pythagoras’ Theorem holds for the triangle with vertices $0, w$ and $z$ with a right angle at $w$. So the locus of points satisfying $\operatorname{Re}((z + w)/(z − w)) = 1$ is the line which passes through $w$ at a right angle to the line segment from 0 to $w$ (and excluding $w$ itself).
#31C Let $a,b,c$ be complex numbers such that $|a| = |b| = |c|$ and $a + b + c = 0$. Show that $a^2 +b^2 +c^2 = 0$.
Proof.
Say $r = |a| = |b| = |c|$ so that $a\bar a=b\bar b=c\bar c=r^2$. If $r=0$ the question is trivial. Otherwise\begin{align*} a^{2}+b^{2}+c^{2} &=(a+b+c)^{2}-2(a b+b c+c a) \\ &=0^{2}-2 a b c\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ &=-2 a b c\left(\frac{\bar{c}}{r^{2}}+\frac{\bar{a}}{r^{2}}+\frac{\bar{b}}{r^{2}}\right) \\ &=-\frac{2 a b c}{r^{2}}(\overline{c+a+b})=0 \end{align*}#32B Let $z_1 = 1+i, z_2 =\sqrt3+i$. Find the real and imaginary parts, modulus and argument of $z_1/z_2$. Deduce$$\cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \quad \text { and } \quad \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$$Solution.
Note\begin{align*}|1+i| &=\sqrt{2} \quad \text { and } \quad \arg (1+i)=\pi / 4 \\|\sqrt{3}+i| &=2 \quad \text { and } \quad \arg (\sqrt{3}+i)=\pi / 6 \end{align*}Hence$$\left|\frac{1+i}{\sqrt{3}+i}\right|=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \quad \text { and } \quad \arg \left(\frac{1+i}{\sqrt{3}+i}\right)=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$$Further$$\frac{1+i}{\sqrt{3}+i}=\left(\frac{1+i}{\sqrt{3}+i}\right)\left(\frac{\sqrt{3}-i}{\sqrt{3}-i}\right)=\frac{(1+\sqrt{3})+i(\sqrt{3}-1)}{4}$$Hence$$\frac{1}{\sqrt{2}} \cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{4} \quad \text { and } \quad \frac{1}{\sqrt{2}} \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{4}$$to give the required expressions. As an alternative approach we also have that
\begin{align*} 2 \cos ^{2} \frac{\pi}{12}-1 &=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \\ & \Longrightarrow \quad \cos \frac{\pi}{12}=\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}=\frac{1}{2} \sqrt{2+\sqrt{3}} \end{align*}taking the positive square root as $\cos (π/12) > 0$. As $(2+\sqrt{3})=\frac{1}{2}(1+\sqrt{3})^{2}$ the two formulae for $\cos (π/12)$ are in agreement.

#33D Let $z \ne 0$. Show there is an integer $1≤n≤628$ such that $−0.01≤\arg(z^n)≤0.01$.
Proof.
We may assume that $|z| = 1$ as the arguments of $z^k$ and $(z/|z|)^k$ are equal. In this case the powers $z^k$ will all lie on the unit circle $|z| = 1$. Note that $6.29 > 2π > 6.28$ and consider splitting the unit circle into 629 arcs associated with the arguments $0≤\arg w < 0.01, 0.01≤\arg w < 0.02,\cdots 6.27≤\arg w < 6.28, 6.28≤\arg w < 2π$.
Suppose for a contradiction that none of the 628 numbers $z, z^2,\cdots , z^{628}$ lie on the arcs $0≤\arg w < 0.01$ or $6.28≤\arg w < 2π$. In this case then two of these 628 numbers must lie on one of the other 627 arcs — say these are $z^i$ and $z^j$ where $628\ge  i > j > 0$. But then $−0.01 <\arg(z^{i−j})=\arg z^i −\arg z^j < 0.01$ and $z^{i−j}$ is amongst the numbers $z,z^2,\cdots,z^{628}$.

#38B By a half-plane of $C$ we mean the set of points that lie to one side of a line. Prove that every half-plane can be written in the form $\Im((z − a)/b) > 0$ for complex numbers $a,b$ where $b \ne 0$.
Proof.
Every half plane can be put in the form $Ax+By > C$ for real numbers $A, B, C$ where $A$ and $B$ aren’t both zero. If $a = a_1 + ia_2, b = b_1 + ib_2$ and $z = x + yi$ then\begin{align*} \operatorname{Im}\left(\frac{z-a}{b}\right) &=\operatorname{Im}\left(\frac{(x+y i)-\left(a_{1}+i a_{2}\right)}{b_{1}+i b_{2}}\right) \\ &=\frac{\operatorname{Im}\left[\left(\left(x-a_{1}\right)+i\left(y-a_{2}\right)\right)\left(b_{1}-i b_{2}\right)\right]}{b_{1}^{2}+b_{2}^{2}} \\ &=\frac{-b_{2}\left(x-a_{1}\right)+b_{1}\left(y-a_{2}\right)}{b_{1}^{2}+b_{2}^{2}} \end{align*}So $\operatorname{Im}\left(\frac{z-a}{b}\right)>0 \Longleftrightarrow-b_{2} x+b_{1} y>b_{1} a_{2}-b_{2} a_{1}$.
So we set $b_1 = B$ and $b_2 = −A$. As $A$ and $B$ aren’t both zero, then $b \ne 0$. If $A\ne 0$ we further set $a_2 = 0$ and $a_1 = C/A$. If $A = 0$ then we set $a_1 = 0$ and $a_2 = C/B$.

#42B Let $\alpha$ be a complex number and $k > 0$. Show the equation $|z−α| + |z +α| = k$ has solutions in $z$ if and only if $2|α|\le k$.
Proof.
For complex $\alpha$ and positive $k$, if there exists a solution $z$ to the equation $|z − α| + |z + α| = k$ then $k = |z + α| + |α − z|\ge |(z + α) + (α − z)| = 2 |α|$.
Conversely suppose $2 |α|≤k$. If $\alpha=0$ then the equation is that of a circle, so suppose $\alpha\ne0$. We shall seek out a solution of the form $z =\lambda\alpha$. Substituting this into the equation we get $(|\lambda − 1| + |\lambda+ 1|)|α| = k$.
We can then see that $λ = k/ (2 |α|) > 1$ solves this equation.