Complex Numbers

Setting up $u$ and $v$.

Write $z=x+\mathrm{i}y$ and substitute into $w=\frac{1+\mathrm{i}z}{\mathrm{i}+z}$. Multiply numerator and denominator by the conjugate of the denominator $(x-\mathrm{i}(1+y))$:$$w=\frac{(1+\mathrm{i}(x+\mathrm{i}y))}{(\mathrm{i}+x+\mathrm{i}y)}=\frac{(1-y)+\mathrm{i}x}{x+\mathrm{i}(1+y)}.$$Multiplying top and bottom by $x-\mathrm{i}(1+y)$:$$w=\frac{[(1-y)+\mathrm{i}x][x-\mathrm{i}(1+y)]}{x^2+(1+y{)}^2}.$$Expanding the numerator: real part is $(1-y)x+x(1+y)=2x$, and imaginary part is $x^2-(1-y)(1+y)=x^2-(1-y^2)=x^2+y^2-1$. Hence$$u=\frac{2x}{x^2+(1+y{)}^2},v=\frac{x^2+y^2-1}{x^2+(1+y{)}^2}.$$Part (i). Set $y=0$, so $u=\frac{2x}{x^2+1}$ and $v=\frac{x^2-1}{x^2+1}$. To see that $w$ lies on the unit circle, compute$$u^2+v^2=\frac{4x^2+(x^2-1{)}^2}{(x^2+1{)}^2}=\frac{(x^2+1{)}^2}{(x^2+1{)}^2}=1.$$So every $z$ on the real axis maps to a point on $u^2+v^2=1$. To parametrise it cleanly, set $x=\tan (\theta /2)$. Then the half-angle formulae give$$u=\frac{2\tan (\theta /2)}{1+{\tan }^2(\theta /2)}=\sin \theta ,v=\frac{{\tan }^2(\theta /2)-1}{{\tan }^2(\theta /2)+1}=-\cos \theta .$$As $x$ ranges over $\mathbb{R}$, $\theta /2$ sweeps $(-\pi /2,\pi /2)$ so $\theta \in (-\pi ,\pi )$, covering the full circle except $\theta =\pm \pi$, i.e. the point $(u,v)=(0,1)$. (One checks directly: $u=0,v=1$ would require $x=0$ and $x^2-1=x^2+1$, impossible.) The locus is the unit circle $u^2+v^2=1$ with the point $(0,1)$ omitted.

Part (ii). Now $y=0$ and $-1<x<1$. From Part (i), $w=(\sin \theta ,-\cos \theta )$ with $x=\tan (\theta /2)$. Since $x\in (-1,1)$ corresponds to $\theta /2\in (-\pi /4,\pi /4)$, i.e. $\theta \in (-\pi /2,\pi /2)$. The locus is the arc of the unit circle $u^2+v^2=1$ with $u\in (-1,1)$ and $v<0$, i.e. the arc in the lower half-plane strictly between $(-1,0)$ and $(1,0)$ (both endpoints excluded).

Part (iii). Now $x=0$, $-1<y<1$. Then $u=0$ and $v=\frac{y^2-1}{(1+y{)}^2}=\frac{(y-1)(y+1)}{(1+y{)}^2}=\frac{y-1}{1+y}$. As $y$ ranges over $(-1,1)$: at $y\to -1^{+}$, $v\to -\infty$; at $y=0$, $v=-1$; at $y\to 1^{-}$, $v\to 0^{-}$. The function $v=\frac{y-1}{1+y}$ is strictly increasing (check: derivative $=\frac{2}{(1+y{)}^2}>0$), so $v$ sweeps $(-\infty ,0)$. The locus is the negative $v$-axis: $u=0$, $-\infty <v<0$.

Part (iv). Set $y=1$, so $u=\frac{2x}{x^2+4}$ and $v=\frac{x^2}{x^2+4}$. Note $v\ge 0$ always, and $u^2+(v-\frac{1}{2}{)}^2=\frac{4x^2+(x^2-4{)}^2/4}{(x^2+4{)}^2}$... let us instead substitute $x=2\tan (\varphi /2)$. Then$$u=\frac{4\tan (\varphi /2)}{4{\tan }^2(\varphi /2)+4}=\frac{\tan (\varphi /2)}{{\tan }^2(\varphi /2)+1}=\frac{1}{2}\sin \varphi ,v=\frac{4{\tan }^2(\varphi /2)}{4({\tan }^2(\varphi /2)+1)}=\frac{1}{2}(1-\cos \varphi ).$$Hence $u^2+(v-\frac{1}{2}{)}^2=\frac{1}{4}{\sin }^2\varphi +\frac{1}{4}{\cos }^2\varphi =\frac{1}{4}$. So $w$ lies on the circle of centre $(0,\frac{1}{2})$ and radius $\frac{1}{2}$. As $x\to \pm \infty$, $\varphi \to \pm \pi$, and both approach $(u,v)=(0,1)$, which is excluded. The locus is the circle $u^2+{\left(v-\frac{1}{2}\right)}^2=\frac{1}{4}$ with the point $(0,1)$ omitted.

Stem. We expand $|z-a{|}^2=(z-a)(z-a{)}^{\ast }$ directly:$$|z-a{|}^2=(z-a)(z^{\ast }-a^{\ast })=z{z}^{\ast }-a{z}^{\ast }-a^{\ast }z+a{a}^{\ast }$$so the given equation is exactly $|z-a{|}^2-r^2=0$, i.e. $|z-a{|}^2=r^2$. This is a circle with centre $a$ and radius $r$. Note that $r^2\ne a{a}^{\ast }$ ensures the circle does not pass through the origin — a fact we shall need.

Part (i). Since $\omega =1/z$ we have $z=1/\omega$, so substitute $z=1/\omega$ and $z^{\ast }=1/{\omega }^{\ast }$ into $|z-a{|}^2=r^2$:$${\left|\frac{1}{\omega }-a\right|}^2=r^2⟹\frac{|1-a\omega {|}^2}{|\omega {|}^2}=r^2⟹|1-a\omega {|}^2=r^2|\omega {|}^2$$Expanding both sides:$$(1-a\omega )(1-a^{\ast }{\omega }^{\ast })=r^2\omega {\omega }^{\ast }$$$$1-a\omega -a^{\ast }{\omega }^{\ast }+a{a}^{\ast }\omega {\omega }^{\ast }=r^2\omega {\omega }^{\ast }$$$$1-a\omega -a^{\ast }{\omega }^{\ast }+(a{a}^{\ast }-r^2)\omega {\omega }^{\ast }=0$$Since $a{a}^{\ast }-r^2\ne 0$, divide through:$$\omega {\omega }^{\ast }-\frac{a}{a{a}^{\ast }-r^2}\omega -\frac{a^{\ast }}{a{a}^{\ast }-r^2}{\omega }^{\ast }+\frac{1}{a{a}^{\ast }-r^2}=0$$Complete the square by writing this as ${\left|\omega -\frac{a^{\ast }}{a{a}^{\ast }-r^2}\right|}^2=\frac{(a^{\ast })(a)}{(a{a}^{\ast }-r^2{)}^2}-\frac{1}{a{a}^{\ast }-r^2}$.

The right-hand side simplifies: $\frac{a{a}^{\ast }}{(a{a}^{\ast }-r^2{)}^2}-\frac{a{a}^{\ast }-r^2}{(a{a}^{\ast }-r^2{)}^2}=\frac{r^2}{(a{a}^{\ast }-r^2{)}^2}$.

So $C^{\prime }$ is a circle with centre $\frac{a^{\ast }}{a{a}^{\ast }-r^2}$ and radius $\frac{r}{|a{a}^{\ast }-r^2|}$.

For $C=C^{\prime }$: the radii must match, giving $(a{a}^{\ast }-r^2{)}^2=1$, i.e. $(|a{|}^2-{\mathbf{r}}^2{)}^2=1$. The centres must also match:$$a=\frac{a^{\ast }}{a{a}^{\ast }-r^2}⟹a(a{a}^{\ast }-r^2)=a^{\ast }⟹a(\pm 1)=a^{\ast }$$If $a{a}^{\ast }-r^2=1$: $a=a^{\ast }$, so $a$ is real. If $a{a}^{\ast }-r^2=-1$: $a=-a^{\ast }$, writing $a=\alpha +i\beta$ gives $\alpha =0$, so $a$ is purely imaginary.

For the sketches: when $a$ is real, the centre lies on the real axis; when $a$ is purely imaginary, it lies on the imaginary axis. In both cases the circle passes close to (but not through) the origin since $r^2\ne |a{|}^2$.

Part (ii). Now $\omega =1/z^{\ast }$, so $z^{\ast }=1/\omega$ and $z=1/{\omega }^{\ast }$. Substituting:$$\frac{1}{{\omega }^{\ast }}\cdot \frac{1}{\omega }-a\cdot \frac{1}{\omega }-a^{\ast }\cdot \frac{1}{{\omega }^{\ast }}+a{a}^{\ast }-r^2=0$$Multiplying through by $\omega {\omega }^{\ast }$ and rearranging as before yields a circle equation with centre $\frac{a}{a{a}^{\ast }-r^2}$. For this locus to equal $C$, we again need $(|a{|}^2-r^2{)}^2=1$, but now the centre condition is $a=\frac{a}{a{a}^{\ast }-r^2}$.

If $a{a}^{\ast }-r^2=1$ this becomes $a=a$, which holds for any $a$ — real, imaginary, or complex. So it is not necessarily the case that $a$ is real or imaginary.

Part (i).

With $w=\frac{2}{z-2}$, compute $w-1$:$$w-1=\frac{2}{z-2}-1=\frac{2-(z-2)}{z-2}=\frac{4-z}{z-2}.$$For $z=3+ti$ (so $z-2=1+ti$):$$w-1=\frac{4-(3+ti)}{1+ti}=\frac{1-ti}{1+ti}.$$Taking the modulus:$$|w-1{|}^2=\frac{1-ti}{1+ti}\cdot \overline{\left(\frac{1-ti}{1+ti}\right)}=\frac{(1-ti)(1+ti)}{(1+ti)(1-ti)}=\frac{1+t^2}{1+t^2}=1,$$so $|w-1|=1$ for all $t\in \mathbb{R}$, independent of $t$. Any $z$ on the line $\mathrm{Re}(z)=3$ has the form $z=3+ti$, so $w$ always satisfies $|w-1|=1$: a circle centred at $1$ with radius $1$.

For the general vertical line $V:\mathrm{Re}(z)=p$ (with $p\ne 2$), write $z=p+ti$. We look for a real centre $c$ such that $|w-c|$ is constant:$$w-c=\frac{2}{z-2}-c=\frac{2-c(z-2)}{z-2}=\frac{(2+2c-cp)-cti}{(p-2)+ti}.$$Computing the squared modulus:$$|w-c{|}^2=\frac{(2+2c-cp{)}^2+c^2{t}^2}{(p-2{)}^2+t^2}.$$For this to be independent of $t$, the coefficients of $t^2$ must be proportional: we need $(2+2c-cp{)}^2=c^2(p-2{)}^2$, i.e.$$(2-c(p-2){)}^2=c^2(p-2{)}^2.$$Expanding: $4-4c(p-2)+c^2(p-2{)}^2=c^2(p-2{)}^2$, giving $4=4c(p-2)$, so$$c=\frac{1}{p-2}.$$The radius is then $r=|c|=\frac{1}{|p-2|}$. The circle has centre $\frac{1}{p-2}$ and radius $\frac{1}{|p-2|}$.

To find when $\mathrm{Im}(w)>0$, write $z-2=(p-2)+ti$:$$w=\frac{2}{(p-2)+ti}=\frac{2((p-2)-ti)}{(p-2{)}^2+t^2},$$so $\mathrm{Im}(w)=\frac{-2t}{(p-2{)}^2+t^2}$. This is positive precisely when $t<0$, i.e. when $\mathrm{Im}(z)<0$.

Part (ii).

For the horizontal line $H:\mathrm{Im}(z)=q$ (with $q\ne 0$), write $z=t+qi$. By analogy with part (i), look for a purely imaginary centre $di$ such that $|w-di|$ is constant:$$w-di=\frac{2-di(z-2)}{z-2}=\frac{(2+dq)+d(2-t)i}{(t-2)+qi}.$$Squared modulus:$$|w-di{|}^2=\frac{(2+dq{)}^2+d^2(2-t{)}^2}{(t-2{)}^2+q^2}=\frac{d^2(t-2{)}^2+(2+dq{)}^2}{(t-2{)}^2+q^2}.$$Independence from $t$ requires $(2+dq{)}^2=d^2{q}^2$, i.e. $4+4dq=0$, giving $d=-\frac{1}{q}$.

The circle has centre $-\frac{i}{q}$ and radius $|d|=\frac{1}{|q|}$.

For the real part: $w=\frac{2(t-2)-2qi}{(t-2{)}^2+q^2}$, so $\mathrm{Re}(w)=\frac{2(t-2)}{(t-2{)}^2+q^2}$. This is positive precisely when $t>2$, i.e. when $\mathrm{Re}(z)>2$.