1. Which of the following quadratic equations require the use of complex numbers to solve them?$$3 x^{2}+2 x-1=0 ; \quad 2 x^{2}-6 x+9=0 ; \quad-4 x^{2}+7 x-9=0$$Solution.
   The discriminants are 16,-36, and -95. So the first and the third require the use of complex numbers to solve them.
2. Show directly that if $zw=0$, where $z, w$ are two complex numbers, then $z=0$ or $w=0$ (or both). Now re-prove this making use of properties of the modulus function.
   Proof.
   Direct Proof: Let $z=x+yi,w=u+vi$, where $x,y,u,v\in\mathbb R$. If $zw=(xu-yv)+(xv+yu)i=0$, then $(xu-yv)^2+(xv+yu)^2=0$, factoring, $(x^2+y^2)(u^2+v^2)=0$, so $z=0\lor w=0$.
   Making use of properties of the modulus function: From $zw=0$, we get $|z|\cdot |w|=|zw|=0$, which means that $|z|=0\lor |w|=0$, so $z=0\lor w=0$.
3. Put each of the following numbers into the form $a+bi$:
   $(3+2i)+(2-7i)=5-5i;\quad(1+2i)(3-i)=5+5i;\quad\frac{1+2i}{3-i}=\frac{1}{10}+\frac{7i}{10};\quad(1+i)^{4}=-4$.
4. Find the modulus and argument of each of the following numbersrac{3\pi}4$
   

   complex number	modulus	argument
   $1+\sqrt{3} i$	2	$\frac\pi3$
   $(2+i)(3-i)$	$5\sqrt{2}$	$\frac\pi2-\arctan7$
   $(1+i)^{5}$	$4\sqrt2$
   $\frac{(1+2 i)^{3} }{(2-i)^{3}}$	1>$-\frac\pi2$

5. Find the two square roots of $-5-12i$. Hence solve the quadratic equation $z^{2}-(4+i) z+(5+5 i)=0$.
   Using $-5-12i=13\left(-\frac5{13}-\frac{12}{13}i\right)$ and
   ngle formula for sines and cosines, we get the two square roots of $-5-12i$ are $\pm(2-3i)$.
   From $z^{2}-(4+i) z+(5+5 i)=0$, letting $z=\frac{w}{2}+2+\frac{i}{2}$, we get $w^2=-5-12i$, so $w=\pm(2-3i)$, so $z=3-i
   1+2i$.
6. On separate Argand diagrams sketch the following sets.
   (i)$|z|<1$; a disk with radius 1 and center origin.
   
   (ii)$\operatorname{Re}z=3$; a line parallel to imaginary axis.
   
   (iii)$|z-1|=|z+i|$; the perpendicular bisector of 1 and $-i$.
   (iv)$\arg (z-i)=\frac\pi4$; the ray with start point $i$ (excluded) whic
   s an angle of $\frac\pi4$ with real axis.
   (v)$-\frac\pi4<\arg z<\frac\pi4$; the area bounded by two rays which form an angle of $\pm\frac\pi4$ with real axis.
   
   (vi)$\operatorname{Re}(z+1)=|z-1|$; a parabola with focus 1 and directrix $
   torname{Re}z=-1$.
   (vii)$|z-3-4 i|=5$; a circle with center $3+4i$ and radius 5.
   
   (viii)$\operatorname{Re}((1+i)z)=1$; a line through 1 and $\frac{1-i}2$.
   
   (ix)$\operatorna
   \left(z^{3}\right)>0$. Let $\theta=\arg z$, the inequality is equivalent to $\sin3\theta>0$. $-\pi<\theta<\pi\Rightarrow-3\pi<3\theta<3\pi$. Therefore $\sin3\theta>0\Leftrightarrow\theta\in\left(-\frac{2\pi}3,-\frac\pi3\right)\cup\left(0,\frac\pi3\right)\cup\left(\frac{2\pi}3,\pi\right)$.
7. Let $z=\operatorname{cis} \theta=e^{i \theta}$ and $n$ be an integer.
   (i) Show that $2 \cos \theta=z+z^{-1}$ and that $2 i \sin \theta=z-z^{-1}$.
   
   (ii) Using De Moivre's theorem, show further that $2 \cos n \theta=z^{n}+z^{-n}$ and that $2 i \sin n \theta=z^{n}-z^{-n}$.
   
   (iii) Deduce that $16 \cos ^{5} \theta=\cos 5 \theta+5 \cos 3 \theta+10 \cos \theta$ and evaluate $\int_{0}^{\pi / 2} \cos ^{5} \theta~\mathrm{d} \theta$.
   
   Solution.
   
   (i) $z=\cos\t
   \sin\theta,z^{-1}=\cos\theta-i\sin\theta$, adding and subtracting, we get $2\cos\theta=z+z^{-1}$ and $2i\sin\theta=z-z^{-1}$.
   (ii) De Moivre's theorem says $z^n=\cos\theta+i\sin\theta,z^{-n}=\cos\theta-i\sin\theta$, adding and subtracting, we get $2\cos n\theta
   ^{-n}$ and $2i\sin n\theta=z^n-z^{-n}$.
   (iii) From (i), we get $32\cos ^{5} \theta=\left(z+z^{-1}\right)^5=\frac{\left(z^2+1\right)^5}{z^5}=z^5+\frac{1}{z^5}+5 z^3+\frac{5}{z^3}+10 z+\frac{10}{z}=2\cos5\theta+10\cos3\theta+20\cos\theta$.
   
   $\int_{0}^{\pi / 2} \cos ^{5} \theta~\mathrm{d} \theta=\frac{1}{16} \left(\frac{1}{5}-\frac{5}{3}+10\right)=\frac{8}{15}$.
8. $\zeta=\operatorname{cis}\frac{2 \pi } 5=e^{2 \pi i / 5}$. Show that $1+\zeta+\zeta^{2}+\zeta^{3}+\zeta^{4}=0$. Show further that $\left(z-\zeta-\zeta^{4}\right)\left(z-\zeta^{2}-\zeta^{3}\right)=z^{2}+z-1$ and deduce that $\cos \frac{2 \pi } 5=\frac{\sqrt{5}-1}4$.
   
   Proof. $\zeta^5=1\Rightarrow(1-\zeta)\left(1+\zeta+\zeta^2+\zeta^3+\zeta^4\right)=0$, but $\zeta\ne1$, so $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$.
   
   $\zeta+\zeta^4+\zeta^2+\zeta^3=-1$, so the coefficient of $z$ term is 1. $\left(\zeta+\zeta^4\right)\left(\zeta^2+\zeta^3\right)=\zeta^3+\zeta^4+\zeta^6+\zeta^7=\zeta^3+\zeta^4+\zeta+\zeta^2=-1$, so the constant term is $-1$.
   
   $\zeta+\zeta^4=e^{2\pi i/5}+e^{8\pi i/5}=e^{2\pi i/5}+e^{-2\pi i/5}=2\cos{\frac{2\pi}5},\zeta^2+\zeta^3=e^{4\pi/5}+e^{6\pi/5}=e^{4\pi/5
   4\pi/5}=2\cos{\frac{4\pi}5}\Rightarrow z=2\cos{\frac{2\pi}5}$ and $z=2\cos{\frac{4\pi}5}$ are 2 roots of quadratic equation $z^2+z-1=0$, but the 2 roots are $\frac{-1\pm\sqrt5}2$, so the positive root $\frac{\sqrt5-1}2$ must be $2\cos\frac{2\pi}5$, so $\cos\frac{2\pi}5=\frac{\sqrt5-1}4$.
9. Write down the seven roots of $z^7+1=0$. By considering the coefficient of $z^{6}$ in the factorization of $z^{7}+1$, show that $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}=\frac{1}{2}$.
   Solution.
   The 7 roots are $\operatorname{cis}\left(\frac\pi7+\frac{2k\pi}7\right),i=0,1,\cdots,6$. By Viète's theorem, the coefficient of $z^{6}$ in $z^7+1$ is $-1+\left(\operatorname{cis}\frac{\pi}7+\operatorname{cis}\frac{13\pi}7\right)+\left(\operatorname{cis}\frac{5\pi}7+\operatorn
   s}\frac{9\pi}7\right)+\left(\operatorname{cis}\frac{3\pi}7+\operatorname{cis}\frac{11\pi}7\right)$
   $=-1+2\cos\frac\pi7+2\cos\frac{5\pi}7+2\cos\frac{3\pi}7$, but this clearly equals 0, so $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}=\frac{1}{2}$.
10. (i) Let $\zeta=\operatorname{cis}\frac{\pi}7=e^{i\pi/7}$. Simplify the expression $\left(\zeta-\zeta^{
    ht)\left(\zeta^{3}-\zeta^{4}\right)\left(\zeta^{5}-\zeta^{2}\right)$. Hence show that $\cos\frac\pi7\cos\frac{3\pi}7\cos\frac{5\pi}7=-\frac18$.
    (ii) Given that $\cos 7 \theta=64 \cos ^{7} \theta-112 \cos ^{5} \theta+56 \cos ^{3} \theta-7 \cos \theta$, rederive the result of (i).
    
    Solution.
    
    (i)$\left(\zeta-\zeta^{6}\right)\left(\zeta^{3}-\zeta^{4}\right)\left(\zeta^{5}-\zeta^{2}\right)=\left(e^{\frac{i \pi }{7}}-e^{\frac{6 i \pi }{7}}\right)\left(e^{\frac{3 i \pi }{7}}-e^{\frac{4 i \pi }{7}}\right) \left(e^{\frac{5 i \pi }{7}}-e^{\frac{2 i \pi }{7}}\right) =-1$.
    
    By $e^{\frac{i \pi }{7}}-e^{\frac{6 i \pi }{7}}=2 \cos\frac{\pi }{7}$,$e^{\frac{3i \pi }{7}}-e^{\frac{4i \pi }{7}}=2 \cos\frac{3\pi }{7}$,$e^{\frac{5\pi }{7}}-e^{\frac{2i \pi }{7}}=2 \cos\frac{5\pi }{7}$, we get $\cos\frac\pi7\cos\frac{3\pi}7\cos\frac{5\pi}7=-\frac18$.
    
    The first step depends on the identity $\left(x+\frac{1}{x}\right) \left(x^2+\frac{1}{x^2}\right) \left(x^4+\frac{1}{x^4}\right)=x^7+\frac{1}{x^7}+x^5+\frac{1}{x^5}+x^3+\frac{1}{x^3
    rac{1}{x}$.
    This identity can be generalized to $\prod_{i=0}^n\left(x^{2^i}+\frac
    {2^i}}\right)=\sum_{i=1}^{2^n}\left(x^{2i-1}+\frac{1}{x^{2i-1}}\right)$.
    (ii)$\cos\frac{2
    (k=0,1,\cdots,6)$ are roots of $64x^7-112x^5+56x^3-7x-1=0$. By Viète's theorem, the product of roots of $64x^7-112x^5+56x^3-7x-1=0$ is $\frac1{64}$. So $\left(\cos\frac\pi7\cos\frac{3\pi}7\cos\frac{5\pi}7\right)^2=\cos\frac{2\pi}7\cos\frac{4\pi}7\cos\frac{6\pi}7\cos\frac{8\pi}7\cos\frac{10\pi}7\cos\frac{12\pi}7=\frac1{64}$. Because $\cos\frac\pi7\cos\frac{3\pi}7\cos\frac{5\pi}7<0,$ we have $\cos\frac\pi7\cos\frac{3\pi}7\cos\frac{5\pi}7=-\frac18$.
    Another way: $\cos\frac{\pi+2k\pi}7(k=0,1,\cdots,6)$ are roots of $64x^7-112x^5+56x^3-7x+1=0$. By Viète's theorem, the product of roots of $64x^7-112x^5+56x^3-7x+1=0$ is $-\frac1{64}$, and $\cos\frac{7\pi}7=-1$, similar to method1, we have $\cos\frac\
    s\frac{3\pi}7\cos\frac{5\pi}7=-\frac18$.
11. (i)Let $a, b, c$ be positive real numbers. By expanding $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ and considering the second factor as a quadratic in $a$, show that $a^{3}+b^{3}+c^{3} \geqslant 3 a b c \quad \text { if } a, b, c>0$.
    (ii) Let $\omega^3=1,\omega\ne1$. Show for any real numbers $a,b,c$ that $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a+\omega b
    a^{2} c\right)\left(a+\omega^{2} b+\omega c\right)$.
    Hence find all real numbers $a,b,c$ which satisfy $a^3+b^3+c^3=3abc$.
    
    Solution.
    
    (i)Expanding, we get $\left(a+\omega b+\omega^{2} c\right)\lef
    mega^{2} b+\omega c\right)=a^2+\omega^{3}b^2+\omega^{3}c^2+(\omega^2+\omega)(ab+bc+ca)=a^2+b^2+c^2-ab-bc-ca$, and $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^3+b^3+c^3-3abc$.
    Because $a^2-ab+
    +c^2-ca=\frac{1}{2} \left((a-b)^2+(c-a)^2+(b-c)^2\right)\geq 0$, we have $a^3+b^3+c^3-3abc\geqslant0$
    In fact, this is the determinant of a circulant matrix, and can be generalised to $n>3$. (ii)$\{(a,a,a)\mid a\in\mathbb R\}\cup\{(a,b,-a-b)\mid a,b\in\mathbb R\}$
12. (i) Let $p,q$ be complex numbers. Show that Viète's substitution-\frac p{3w}$ turns the equation $z^{3}+p z=q$ into the quadratic in $w^3$.
    (ii)Use Viète's substitution to solve the cubic $z^3-12z+8=0$.
    
    Solution.
    
    (i)$\left(w-\frac{p}{3 w}\right)^3+p \left(w-\frac{p}{3 w}\right)-q=-\frac{p^3+27 q w^3-27 w^6}{27 w^3}$, so we get a quadratic in $w^3$: $\frac{p^3}{27}+q w^3-(w^
    $
    (ii)Let $p=-12,q=-8$, we get $w^6+8 w^3+64=0$, so $w^3=8\operatorname{cis}\frac{\pm2\pi}3$, so $w=2\operatorname{cis}\frac{\pm2k\pi}9\;\;(k=1,2,4)$
    
    so $z=2\operator
    is}\frac{2k\pi}9+2\operatorname{cis}\frac{2k\pi}9=4\cos\frac{2k\pi}9\;\;(k=1,2,4)$
13. (i) Let $P(z)$ be a polynomial with (possibly repeated) roots $\alpha_1,\alpha_2,\cdots,\alpha_k$. Show that $\frac{P^{\prime}(z)}{P(z)}=\frac{1}{z-\alpha_{1}}+\frac{1}{z-\alpha_{2}}+\cdots+\frac{1}{z-\alpha_{k}}$.
    (ii) Deduce that if $\operatorname{Im} \alpha_{i}>0$ for each $i$, then $\operatorname{Im} \beta>0$ for any root $\beta$ of $P'(z)$.
    
    (iii) (Gauss–Lucas theorem) If $P$ is a (nonconstant) polynomial with complex coefficients, all zeros of $P'$ belon
    he convex hull of the set of zeros of $P$.
    (iv) Deduce further that if all the roots $\alpha$ of a polynomial $P(z)$ satisfy $|\alpha|<R$ then all the roots $\beta$ of $P'(z)$ satisfy $|\beta|<R$.
    
    Solution.
    
    (i)Let $P(z)=a(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_k)$, using product rule and induction, we have $\displaystyle P'(z)=a\sum_{i=1}^k\prod_{\substack{1\le j\le k\\j\ne i}}(z-z_j)$.<
    Therefore $\frac{P^{\prime}(z)}{P(z)}=\frac{1}{z-\alpha_{1}}+\frac{1}{z-\alpha_{2}}+\cdots+\frac{1}{z-\alpha_{k}}$.
    
    Another way: Taking logarithm of both sides, $\log P(z)=\log(z-\alpha_1)+\log(z-\alpha_2)+\cdots+\log(z-\alpha_k)$, differentiating, $\fr
    z)}{P(z)}=\frac1{z-\alpha_1}+\frac1{z-\alpha_2}+\cdots+\frac1{z-\alpha_k}$.
    (ii)If $z$ is also a root of $P(z)$, so $z$ is among $\alpha_{i}$, we deduce that $\operatorname{Im}z>0$.
    
    If $z$ is no
    ot of $P(z)$, we have $\frac{P'(z)}{P(z)}\ne0\Leftrightarrow P'(z)\ne0$. We prove the contrapositive: If $\operatorname{Im}z\le0$, then $z$ is not a root of $P'(z)$.
    In fact, if $\operatorname{Im}z\le0$, we have $\operatorname{Im}(z-\alpha_k)<0\Rightarrow\operatorname{Im}\frac{1}{z-\alpha_k}>0\forall k\Rightarrow\operatorname{Im}\frac{P'(z)}{P(z)}>0\Rightarrow z$ is not a root of $P'(z)$.
    
    (iii)We prove the equivalent statement: if all the roots of a polynomial $P$ lie in a half-plane then so do all the roots of $P'$.
    
    If all the roots of a polynomial lie in half-plane $\Im((z − a) /b) > 0$ (see #38) then we can make a change of variable $w = (z − a) /b$. In terms of the new coordinate $w$ the plane is now the region $\Im w > 0$ whilst the polynomial $P(z)$ equals $P (bw + a)$ which remains a polynomial (in $w$) and we can apply the previous part.
    
    Another way:
    
    If $z$ is a zero
    '$ and $P$, the theorem is trivial; if $z$ is a zero of $P^\prime$ and $P(z) \neq 0$, then$$\sum_{i=1}^n \frac{1}{z-\alpha_i}=0\Leftrightarrow\sum_{i=1}^n \frac{\overline{z}-\overline{\alpha_i}} {|z-\alpha_i|^2}=0\Leftrightarrow\overline{z}= \left(\sum_{i=1}^n\frac{1}{|z-\alpha_i|^2}\overline{\alpha_i}\right)/\left(\sum_{i=1}^n \frac{1}{|z-\alpha_i|^2}\right)$$We see that $z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $\alpha_i$.
    (iv) Let $w$ be an arbitrary point on the circle $|z| = R$. If the roots of the polynomial $P(z)$ lie in the disc $|z| < R$, then in particular the roots lie in the half-plane which contains the disc, and whose bounding line is tangential to the circle $|z| = R$ at the point $w$. Hence, by Lucas’ Theorem, all the roots of $P'(z)$ lie in this half-plane.
    
    As we vary $w$ around the circle $|z| = R$, then we see that all the roots of $P'(z)$ lie in the intersection of all these half-planes. But that intersection is precisely the disc $|z| < R$.
14. (i)Show, for any complex number $z$ and p
    e integer $n$, that$$z^{2 n}-1=\left(z^{2}-1\right) \prod_{k=1}^{n-1}\left\{z^{2}-2 z \cos\frac{k \pi}{n}+1\right\}$$(ii) Deduce that for any real $\theta$ that $\displaystyle\sin n \theta=2^{n-1} \sin \theta \prod_{k=1}^{n-1}\left\{\cos \theta-\cos\frac{k \pi}{n}\right\}$.
    (iii) Determine $\displaystyle\prod_{k=1}^{2 n} \cos\frac{k \pi}{2 n+1}$.
    
    Solution.
    
    (i)Because $\operatorname{cis}\frac{k\pi}n+\operatorname{cis}\frac{-k\pi}n=2\cos\frac{k\pi}n$ and $\operatorname{cis}\frac{k\pi}n\operatorname{cis}\frac{-k\pi}n=1$, we have$$\displaystyle z^{2 n}-1=\prod_{k=-n+1}^{n}\lef
    peratorname{cis}\frac{2k\pi}{2n}\right)=\left(z^2-1\right)\prod_{k=1}^{n-1}\left(z-\operatorname{cis}\frac{k\pi}{n}\right)\left(z-\operatorname{cis}\frac{-k\pi}{n}\right)=\left(z^{2}-1\right) \prod_{k=1}^{n-1}\left\{z^{2}-2 z \cos\frac{k \pi}{n}+1\right\}$$(ii)Dividing (i) by $z^n$, we get $z^{n}-z^{-n}=\left(z-\frac1z\right)\prod_{k=1}^{n-1}\left\{z+\frac1z-2\cos\frac{k \pi}{n}\right\}$.
    Let $z=\opera
    e{cis}\theta$, we get $2i\sin{n\theta}=2i\sin\theta\prod_{k=1}^{n-1}\left\{2\cos\theta-2\cos\frac{k\pi}{n}\right\}\Rightarrow\sin{n\theta}=2^{n-1}\sin\theta\prod_{k=1}^{n-1}\left\{\cos\theta-\cos\frac{k\pi}{n}\right\}$.
    (iii)Proof One. By (ii), $\displaystyle\sin\frac{(2n+1)\pi}2=2^{2n}\sin\frac\pi2\prod_{k=1}^{2n}\left\{\cos\frac\pi2-
    rac{k\pi}{2n+1}\right\}\Rightarrow\prod_{k=1}^{2n}\cos\frac{k\pi}{2n+1}=\frac{(-1)^n}{2^{2n}}$
    Proof Two. Let $S=\sin \frac{2\pi}{2n+1}\sin \frac{4\pi}{2n+1}\sin \frac{6\pi}{2n+1} \cdots \sin \frac{2(2n-1)\pi}{2n
    n \frac{2(2n)\pi}{2n+1}$ and $C=\prod_{k=0}^{2n}\cos\frac{2k\pi}{2n+1}=\prod_{k=1}^{2n}\cos\frac{2k\pi}{2n+1}$.
    Then $2^{
    dot S \cdot C = \sin \frac{4\pi}{2n+1}\sin \frac{8\pi}{2n+1}\sin \frac{12\pi}{2n+1} \cdots \sin \frac{4n\pi}{2n+1}\sin \frac{4(n+1)\pi}{2n+1} \cdots \sin \frac{4(2n-1)\pi}{2n+1}\sin \frac{4(2n)\pi}{2n+1}$.
    We calculate:
    
    $\sin \frac{4(n+1)\pi}{2n+1}=\sin \frac{\pi(4n+2+2)}{2n+1}=\sin\left(2\pi+ \frac{2\pi}{2n+1}\right)=\sin \frac{2\pi}{2n+1}$.
    
    $\sin \frac
    )\pi}{2n+1}=\sin \frac{\pi(4n+2+6)}{2n+1}=\sin\left(2\pi+ \frac{6\pi}{2n+1}\right)=\sin \frac{6\pi}{2n+1}$.
    $\cdots$.
    
    $\sin \frac{4(2n-1)\pi}{2n+1}=\sin \frac{\pi(4n+2+4n-6)}{2n+1}=\si
    (2\pi+ \frac{(4n-6)\pi}{2n+1}\right)=\sin \frac{2(2n-3)\pi}{2n+1}$.
    $\sin \frac{4(2n)\pi}{2n+1}=\sin \frac{\pi(
    n-2)}{2n+1}=\sin\left(2\pi+ \frac{(4n-2)\pi}{2n+1}\right)=\sin \frac{2(2n-1)\pi}{2n+1}$.
    Conclusion:
    
    $2^{2n}S \cdot C
    ghtarrow C=\frac{1}{2^{2n}}\Rightarrow\displaystyle\prod_{k=1}^{2 n} \cos\frac{k \pi}{2 n+1}=\frac{(-1)^n}{2^{2n}}$
15. Let $k>0$.
    i) Sketch the curve $C_k$ with equation $\left|z+\frac1z\right|=k$
    
    ii) What are the extrema of $|z|$ on $C_k$?
    
    iii) Show that the curve $C_2$ consists of precisely two circles.
    
    Solution.
    
    i) Let $z=x+iy$, where $x,y\in\mathbb R$. We have $\left|z+\frac1z\right|=k\Leftrightarrow\left(\frac{x}{x^2+y^2}+x\right)^2+\left(y-\frac{y}{x^2+y^2}\right)^2=k^2$.
    
    Rearrange, we get $\left(x^2 + y^2\right)^2=\left(k^2-2\right)x^2+\left(k^2+2\right)y^2-1$. This is a spiric sectionch is a special kind of toric section.
    Note that $\left|z+\frac{1}{z}\right|^{2}=\left(z+\frac{1}{z}\right)\left(\bar{z}+\frac{1}{\bar{z}}\right)=z \bar{z}+\frac{z}{\bar{z}}+\frac{\bar{z}}{z}+\frac{1}{z \bar{z}}$.
    
    If $z=r \operatorname{cis} \theta$, then\begin{align*}\left|z+\frac{1}{z}\right|^{2} &=r^{2}+\frac{r \operatorname{cis} \theta}{r \operatorna
    }(-\theta)}+\frac{r \operatorname{cis}(-\theta)}{r \operatorname{cis} \theta}+\frac{1}{r^{2}} \\ &=r^{2}+\frac{1}{r^{2}}+\operatorname{cis}(2 \theta)+\operatorname{cis}(-2 \theta) \\ &=r^{2}+\frac{1}{r^{2}}+2 \cos 2 \theta \end{align*}So we have$$\left|z+\frac{1}{z}\right|=k \Longleftrightarrow r^{2}+\frac{1}{r^{2}}=k^{2}-2 \cos 2 \theta$$Consider now the graph of $y = x +1/x$ for $x = r^2 > 0$. This has minimum at $x = 1, y = 2$. As $−1≤\cos 2θ≤1$, then $k^{2}-2 \cos 2 \theta>2$ for at least some values of $\theta$, and the curve is non-empty. We provide sketches:
    Error! Click to view log.On this curve we see $|z| = r$ is maximal when $k^2 − 2\cos 2\theta$ is maximal, which is when $\cos 2\theta = −1$ at $\theta = ±π/2$. For these values of $\theta$ we have $r^2+\frac1{r^2}=k^2+2⇒$$\left(r-\frac{1}{r}\right)^{2}=k^{2}$⇒$r-\frac{1}{r}=k$⇒$r^{2}-k r-1=0$⇒$r=\frac{k+\sqrt{k^{2}+4}}{2}$.
    ii) Let $z=re^{i\theta}$, then $\left|r+\frac1r\right|=k\Rightarrow r^2+\frac1{r^2}+2\cos2\theta=k^2\Rightarrow \left(r+\frac1r\right)^2=k^2-4\sin^2\theta$. [$k\ge 2$ since $r+\frac1r\ge2$.]
    
    Now, differentiating both sides with respect to $\theta$ yields $\left(2r-\frac{2}{r^3}\
    \frac{dr}{d\theta}=-4\sin2\theta$ whereupon setting $\frac{dr}{d\theta}$ to $0$ reveals that $2\theta = \ell \pi$ for some integer $\ell$. So the extrema for $r$ occurs when $\theta =0$ or $\frac\pi2$.
    The minimum is $\sqrt{\frac{2+k^2-k\sqrt{k^2+4}}{2}}$, the maximum is $\sqrt{\frac{2+k^2+k\sqrt{k^2+4}}{2}}$, both occur when $\theta=\frac\pi2$.
    iii) Let $z=\rho e^{i\theta}$, then the angle between $z$ and $\frac1z$ is $2\theta$, so $k^2=\rho^2+\frac1{\rho^2}+2\cos{2\theta}=4$.
    $\rho^2+\frac1{\rho^2}=4-2\left(\cos^2\theta-\sin^2\theta\right)$, multiplying by $\rho^2$, we get $\left(x^2+y^2\right)^2+1=2x^2+6y^2$, factoring, $\left(x^2+y^2+2y-1\right)\left(x^2+y^2-2y-1\right)=0$.
    So the curve $C_2$ is the union of two circles with radius $\sqrt2$, centers $(0,\pm1)$.