Express the following complex numbers in the form \(x + i y\) with \(x,y\) real:
(a) \(\frac{(1+2i)(5+2i)}{(2-3i)(1-i)},\) (b)
\(\frac{1+4i}{1-i} +
\frac{1-4i}{1+i}.\)
(a) \(\frac{(1+2i)(5+2i)}{(2-3i)(1-i)} =
\frac{1+12i}{-1-5i} = \frac{(1+12i)(-1-5i)}{(-1-5i)(-1+5i)} =
\frac{-61-7i}{26} = -\frac{61}{26} - \frac{7}{26}i\).
(b) \(\frac{1+4i}{1-i} + \frac{1-4i}{1+i} =
\frac{(1+4i)(1+i)+ (1-4i)(1-i)}{(1-i)(1+i)} = \frac{(-3+5i) +
(-3-5i)}{2} = \frac{-6}{2} = -3.\)
Evaluate the product \((1+i)i(1+i)\), first in the “usual algebraic way", then by writing \(i\) and \(1+i\) in polar form.
We have \((1+i)i(1+i)=(i-1)(1+i) =-2\). On the other hand \(|1+i|=\sqrt 2\) so \(1+i = \sqrt 2 e^{i\pi/4}\), and \(i=e^{i\pi/2}\). Thus, the required product has modulus \(\sqrt 2\ . 1\ . \sqrt 2=2\) and argument \(\pi/4+ \pi/2+\pi/4=\pi\). The product is therefore \(2 e^{i\pi} = -2\), as before.
Write the following complex numbers in polar form:
(a) \(2-i2\sqrt 3,\) (b) \(\frac{1}{2+i} - \frac{1}{2-i} ,\) (c) \(\frac{2-i2\sqrt 3}{1+i}.\)
(a) \(|2-i2\sqrt 3| = \sqrt{4+12} =
4\), so we have \(2-i2\sqrt 3 =
4(\frac{1}{2} - i\frac{\sqrt 3}{2}) = 4e^{i5\pi/3}.\)
(b) \(\frac{1}{2+i} - \frac{1}{2-i} =
-\frac{2i}{5} = \frac{2}{5}e^{-i\pi/2}.\)
(c) \(1+i = \sqrt2 e^{i\pi/4}\), so we
have \(\frac{1}{1+i} = \frac{1}{\sqrt2}
e^{-i\pi/4}\). Thus, \(\frac{2-i2\sqrt
3}{1+i} = 4e^{i5\pi/3} \cdot \frac{1}{\sqrt2} e^{-i\pi/4} = 2\sqrt2
e^{i17\pi/12}.\)
Find all complex numbers \(z\) for
which:
(a) \(|\mathop{\text{Re}}(z)| = |z|\),
(b) \(\mathop{\text{Im}}(z) = |z|\),
(c) \(|z|^2 = z^2\).
(a) If \(z = x+iy\), then the
equation reads \(|x|= \sqrt{x^2 +
y^2}\). This is true if and only if \(y=0\) (or equivalently if \(z\) is real).
(b) We have \(y= \sqrt{x^2 + y^2}\).
This is true if and only if \(x=0\) and
\(y \geq 0\) (because the real square
root function is always non-negative). Equivalently, we must have that
\(z\) is purely imaginary with \(Im(z) \geq 0\).
(c) The equation clearly holds for \(z=0\). For \(z
\ne 0\), we that that \(|z|^2=z\bar{z}=z^2\) is equivalent to \(z=\bar{z}\), which means \(z\) real. So the equation holds if and only
if \(z\) is real. For example \(i^2=-1\), but \(|i|=1\).
If \(w = \frac{z-1}{z+1}\) show that \(\mathop{\text{Re}}(w) = \frac{|z|^2-1}{|z|^2+2Re(z) +1}\) and \(\mathop{\text{Im}}(w) = \frac{2Im(z)}{|z|^2+2Re(z) +1}.\)
\(w = \frac{z-1}{z+1} = \frac{(z-1)(\bar{z} + 1)}{(z+1)(\bar{z} + 1)} = \frac{|z|^2 + z - \bar{z}-1 }{|z|^2+z + \bar{z} +1} = \frac{|z|^2-1 + 2Im(z)i}{|z|^2+2Re(z) +1}.\)
Show that (a) \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\). (b) \(\overline{z_1z_2}=\overline{z_1}\ \overline{z_2}\).
[You could prove these directly by using the definition of conjugation, but there are nicer ways: For (a) note complex conjugation viewed as a reflection in \(\mathbb R^2\) is a linear map. For (b), you could use the fact that \(\overline{z_1 z_2}(z_1z_2) = |z_1z_2|^2\).]
The “nice" ways go as follows: (a) The map \(z \mapsto \bar{z}\) when viewed as a map from \(\mathbb R^2\) to \(\mathbb R^2\) corresponds to \((x,y) \mapsto (x,-y)\). But this map is clearly linear and hence \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\). For (b), we first note the claim holds for if either \(z_1\) or \(z_2\) are zero. Otherwise, we can divide \(\overline{z_1z_2}z_1z_2 = |z_1z_2|^2=|z_1|^2|z_2|^2 = z_1\bar{z_1}z_2\bar{z_2}\) by \(z_1z_2 \ne 0\) and obtain the claim.
Otherwise: (a) Let \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\). Then \[\overline{z_1}+\overline{z_2}= x_1-iy_1+x_2-iy_2=x_1+x_2-i(y_1+y_2),\] while \[\overline{z_1+z_2}= \overline{x_1+iy_1+x_2+iy_2}=\overline{x_1+x_2+i(y_1+y_2)}=x_1+x_2-i(y_1+y_2).\] (b) \[\overline{z_1}\ \overline{z_2}=(x_1-iy_1)(x_2-iy_2)=x_1x_2-y_1y_2-i(x_1y_2+x_2y_1),\] while \[\overline{z_1z_2}=\overline{(x_1+iy_1)(x_2+iy_2)}=\overline{x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)}=x_1x_2-y_1y_2-i(x_1y_2+x_2y_1).\]
For each pair of complex numbers \(z_1\) and \(z_2\) prove the parallelogram identity: \[|z_1 + z_2|^2 + |z_1-z_2|^2 = 2 \bigl( |z_1|^2 + |z_2|^2 \bigr).\] Interpret this equation geometrically.
\[\hbox{LHS}=(z_1 + z_2)(\overline{z}_1 + \overline{z}_2)+(z_1 - z_2)(\overline{z}_1 - \overline{z}_2)=2(z_1\overline{z}_1+z_2\overline{z}_2)= \hbox{RHS}.\] The equation tells you that the sum of the square of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.
What is the geometric meaning of the following functions \(f(z)\) as transformations \(z\mapsto f(z)\) from \(\mathbb C\) to \(\mathbb C\)?
(a)\(\> f(z) = 2z, \qquad \> {\rm
(b)}\> f(z) = -z, \qquad\> {\rm (c)}\> f(z) =
(1+i)z,\)
(d)\(\> f(z) = -\bar{z}, \qquad{\rm (e)}
\>f(z) = z/|z|, \quad\>\>\, {\rm (f)}\> f(z) = 1-i+z,
\qquad\>\>{\rm (g)}\> f(z) =1-i+(1+i)z\).
(a) Radial expansion away from the origin by a factor of 2.
(b) Reflection in the origin (or, rotation about the origin by \(\pi\) degrees since \(-1=e^{i\pi}\)).
(c) \((1+i)=\sqrt 2 e^{i\pi/4}\), so
rotation about origin through angle \(\pi/4\) anticlockwise, followed by a radial
expansion away from the origin by a factor of \(\sqrt 2\).
(d) A point \(z=x+iy\) goes to \(-\bar z = -x+iy\), so reflection in the
imaginary axis.
(e) Radial projection onto the unit circle.
(f) Translation through \(1-i\).
(g) This is (c) followed by (f); i.e., a rotation about origin through
angle \(\pi/4\) anticlockwise, followed
by a radial expansion away from the origin by a factor of \(\sqrt 2\), followed by translation through
\(1-i\).
Draw the following sets of points in the complex plane. \[(a) \ \ z+\bar z=2\ , \quad\qquad (b) \ \ z-\bar z=3i\ , \qquad\quad (c) \ \ {\left|\bar z\right|}=1\ ,\qquad \quad (d) \ \ {\left|z-i\right|}=1\ .\]
(a) Let \(z=x+iy\). Then \(2=z+\bar z=2x\), so the required set is the
line \(x=1\), parallel to the imaginary
axis. (In other words, the set of complex numbers \(z\) with \(Re(z)
= 1\).)
(b) Let \(z=x+iy\). Then \(3i= z-\bar z=2yi\), so the required set is
those points with imaginary part \(y=3/2\); i.e., the line through \(3i/2\), parallel to the real axis.
(c) We have \(|\bar z| = |z|\), and
\(|z|\) is the distance of \(z\) from the origin \(0\), so \(|\bar
z|=1\) is the circle of radius \(1\) centre \(0\).
(d) This is the circle of radius 1 with centre \(i\) (so this circle passes through 0
tangential to the real axis).
Solve the following equations in complex numbers, and mark the
solutions on a picture of the complex plane:
(a) \(|z+2|=|z-2|\), (b) \(\bar z = 1/z,\) (c) \(z =\frac{
\mathop{\text{Re}}z+\mathop{\text{Im}}z}{2},\) (d) \(|(z-2 )(\bar z-2)| = 1.\)
The pictures/sketches are not displayed in this solution sheet.
(a) \(|z+2|=|z-2| \Leftrightarrow
|z+2|^2=|z-2|^2 \Leftrightarrow (z+2)(\overline {z}+2)=(z-2)(\overline
{z}-2) \Leftrightarrow 2(z+\overline {z})=2( -z-\overline
{z})\Leftrightarrow z+\overline {z}=0 \Leftrightarrow z\) is
purely imaginary. This is clear geometrically because the imaginary axis
is the set of point in the complex plane equidistant from \(2\) and \(-2\).
(b) We must have \(z\neq 0\). Then,
\(\bar z = 1/z \Leftrightarrow \bar z
z=1 \Leftrightarrow |z|=1\), so \(\bar
z = 1/z\) if and only if \(z\)
lies on the unit circle centre the origin.
(c) Let \(z=x+iy\). Then \(\mathop{\text{Re}}z+\mathop{\text{Im}}z=x+y\),
so that \(z =\frac{
\mathop{\text{Re}}z+\mathop{\text{Im}}z}{2}\) if and only if
\(z=0\).
(d) \(|(z-2)(\bar z-2)|
=|(z-2)\overline{(z-2)}| = |z-2||\overline{ z-2}| = |z-2|^2\), so
the equation simplifies to \(|z-2| =
1\), and the solutions lie on the circle centred at \(2\) with radius 1.
What do the following equations represent geometrically? Give
sketches.
(i) \(|z+2|=6\) (ii) \(|z-3i|=|z+i|\) (iii) \(|iz-1|=|iz+1|\) (iv) \(|z+1-i|=|\bar z- 1-i|\).
(i) \(|z+2|= |z-(-2)|\), which is
the distance of \(z\) from \(-2\), so \(|z+2|=6\) is the circle radius 6 centre
\(-2\).
(ii) The equation represents those points with equal distance to \(3i\) and \(-i\). Thus, the locus is the perpendicular
bisector of the line segment joining \(3i\) and \(-i\); that is, the line parallel to the
real axis passing through \(i\).
Alternatively, for \(z = x + iy\) one
could notice \(|z-3i|=|z+i| \Leftrightarrow
|z-3i|^2=|z+i|^2 \Leftrightarrow x^2 + (y-3)^2= x^2 + (y-1)^2\).
Simplifying, we have that this is equivalent to \(9-6y = 2y+1\) and in turn \(y=1\).
(iii) \(|iz-1|=|iz+1| \iff |i(z+i)|=|i(z-i)|
\iff |(z+i)|=|(z-i)|\). This is the perpendicular bisector of
the line segment joining \(-i\) and
\(i\); in other words, the real
axis.
(iv) \(|z+1-i|=|\bar
z-1-i|\iff |z-(-1+i)|=|z-(1-i)|\). The perpendicular bisector of
\(-1+i\) and \(1-i\) is the line \(y=x\). Alternatively, notice we must have
\((x+1)^2 + (y-1)^2 = (x-1)^2 +
(-y-1)^2\), which simplifies to \(y=x\).
Describe geometrically the subsets of \(\mathbb C\) specified by
(i) \(\mathop{\text{Im}}(z+i)>2\)
(ii) \(1<\mathop{\text{Re}}z\le 2\)
(iii) \(|z-1-i|>1\)
(iv) \(|z-1+i|\ge |z-1-i|\) (v) \(|z+2-i| < |iz-1+2i|\) (vi) \(1< |z-1|<2\).
(i) If \(z=x+iy\), then \(\mathop{\text{Im}}(z+i)>2 \Leftrightarrow y+1
> 2 \Leftrightarrow y>1\). Thus, the subset is the area of
the plane strictly above the line \(y=1\).
(ii) This is the vertical strip in the \((x,y)\)-plane between the lines \(x=1\) and \(x=2\). The line on the left not included
whereas the one on the right is included.
(iii) \(|z-1-i| = |z-(1+i)|\) so the
equation represents those points \(z\)
whose distance to \(1+i\) is strictly
greater than \(1\). Thus, the equation
represents the region outside the circle with centre \(1+i\) and radius \(1\).
(iv) \(|z-1+i| = |z-(1-i)|\) and \(|z-1-i| = |z-(1+i)|\), so the equation
represents the points \(z\) closer to
\(1+i\) than to \(1-i\). This is precisely the upper half
place (including the real axis).
(v) \(|z+2-i|= |z-(-2+i)|\) and \(|iz-1+2i| = |i| |z-\frac {1}{i}+2| = |z+ i+2| =
|z-(-2-i)|\). Thus the solutions are those points strictly closer
to \(-2+i\) than to \(-2-i\). This is precisely the upper half
plane (with the real axis excluded).
(vi) The equation represents those points whose distance to \(1\) is between \(1\) and \(2\), so the region (called an ‘annulus’)
inside the circle centred at \(1\) of
radius \(2\) and outside the circle
centred at \(1\) of radius \(1\).
Show that the equation \(|z-a|= \lambda |z-b|\), where \(a\) and \(b\) are complex numbers and \(\lambda>0\), describes a circle in the complex plane if \(\lambda \neq 1\). [In fact, every circle in the complex plane can be written in this form!] What geometric figure is represented when \(\lambda = 1\)?
Put \(z=x+iy\) as usual, and let \(a=a_1 + i a_2\) and \(b=b_1+ib_2\). Then \((x-a_1)^2 + (y-a_2)^2 = \lambda^2(x-b_1)^2 + \lambda^2(y-b_2)^2\). Grouping the terms and factorising we find that this ‘simplifies’ to \[\left( x- \frac{a_1-\lambda^2b_1}{1-\lambda^2} \right)^2 + \left( y- \frac{a_2-\lambda^2b_2}{1-\lambda^2} \right)^2 = \frac{\lambda^2}{1-\lambda^2} \left( (a_1-b_1)^2 + (a_2-b_2)^2\right).\] Denoting the terms in the above by \((x-c_1)^2 + (y-c_2)^2 = r^2\) we see that \(|z-c_1-ic_2| = r\) and the equation represents a circle centred at \(c_1 + i c_2\) of radius \(r>0\).
(i) Apply induction to show De Moivre’s formula: \((\cos(x)+i\sin(x))^n = \cos(nx) + i
\sin(nx)\).
(ii) Use this to write \(\cos(3x)\) as
a polynomial in \(\cos(x)\); namely
show that \(\cos(3x) = 4\cos^3(x) -
3\cos(x)\).
(i) When \(n=1\) the statement is
trivial. Assume it holds for \(n=k\),
then \[(\cos(x)+i\sin(x))^{k+1} =
(\cos(x)+i\sin(x))^k(\cos(x)+i\sin(x)) =
(\cos(kx)+i\sin(kx))(\cos(x)+i\sin(x))\] Expanding gives \[(\cos(kx)\cos(x) - \sin(kx)\sin(x)) + i(\sin(kx)
\cos(x) + \sin(x) \cos(kx)) = \cos((k+1)x) + i \sin((k+1)x),\] by
the real double angle formula. Note that in other words we have \(\left[e^{ix}\right]^n = (\cos(x)+i\sin(x))^n =
\cos(nx) + i \sin(nx) = e^{inx}\).
(ii) Considering the above for \(n=3\),
observe that \(\cos(3x) =
\mathop{\text{Re}}\left((\cos(x)+i\sin(x))^3 \right)\).
Expanding, we have \[(\cos(x)+i\sin(x))^3 =
(\cos^3(x)-3\cos(x)\sin^2(x)) + i(3\cos^2(x)\sin(x)-\sin^3(x)).\]
Finally, \(\cos(3x) =
\cos^3(x)-3\cos(x)\sin^2(x) = \cos^3(x)-3\cos(x)(1-\cos^2(x)) =
4\cos^3(x) - 3\cos(x)\).
Write \((1+i\sqrt 3)^{100}\) in \(x+iy\) form.
We have \((1+i\sqrt 3)=2e^{i\pi/3}\), so by De Moivre \((1+i\sqrt 3)^{100}=2^{100}e^{100i\pi/3}= 2^{100}e^{99i\pi/3 + i\pi/3}= 2^{100}e^{33i \pi +i\pi/3}= -2^{100}e^{i\pi/3} =- 2^{99}(1+i\sqrt 3)\).
Show that the inverse of the stereographic projection \(P:\mathbb{S}^{2}\setminus\{N\}\to\mathbb C\) is given by \[P^{-1}\left( z\right)=\left( \frac{2\mathop{\text{Re}}\left( z\right)}{1+{\left|z\right|}^2},\frac{2\mathop{\text{Im}}\left( z\right)}{1+{\left|z\right|}^2},\frac{{\left|z\right|}^2-1}{1+{\left|z\right|}^2}\right).\]
Recall that we have that \[\mathop{\text{Re}}\left( z\right)+i\mathop{\text{Im}}\left( z\right) = P(\xi,\eta,\zeta)=\frac{\xi}{1-\zeta}+i\frac{\eta}{1-\zeta}\] and we would like to express \(\left( \xi,\eta,\zeta\right)\in \mathbb{S}^{2}\setminus\{N\}\) with \(\mathop{\text{Re}}\left( z\right)\) and \(\mathop{\text{Im}}\left( z\right)\). We have that \[\mathop{\text{Re}}\left( z\right) = \frac{\xi}{1-\zeta},\qquad \mathop{\text{Im}}\left( z\right)= \frac{\eta}{1-\zeta}\] or \[\xi = \left( 1-\zeta\right)\mathop{\text{Re}}\left( z\right),\qquad \eta=\left( 1-\zeta\right)\mathop{\text{Im}}\left( z\right).\] Since \(\left( \xi,\eta,\zeta\right)\in \mathbb{S}^{2}\setminus\{N\}\) we have that \[1=\xi^2 + \eta^2 + \zeta^2 = \left( 1-\zeta\right)^2 \mathop{\text{Re}}\left( z\right)^2+\left( 1-\zeta\right)^2 \mathop{\text{Im}}\left( z\right)^2 +\zeta^2 = \left( 1-\zeta\right)^2 {\left|z\right|}^2 + \zeta^2\] We can rewrite the above as \[\left( 1+{\left|z\right|}^2\right)\zeta^2 - 2{\left|z\right|}^2\zeta + {\left|z\right|}^2-1=0\] whose solutions are \[\zeta_{1,2} = \frac{2{\left|z\right|}^2\pm \sqrt{4{\left|z\right|}^4 - 4\left( 1+{\left|z\right|}^2\right)\left( {\left|z\right|}^2-1\right)}}{2\left( 1+{\left|z\right|}^2\right)} = \frac{{\left|z\right|}^2 \pm 1}{1+{\left|z\right|}^2}.\] Since \(\zeta\not=1\) we are only left with the option \[\zeta = \frac{{\left|z\right|}^2-1}{1+{\left|z\right|}^2}.\] To find \(\xi\) and \(\eta\) we notice that \[1-\zeta = 1-\frac{{\left|z\right|}^2-1}{1+{\left|z\right|}^2}=\frac{2}{1+{\left|z\right|}^2}.\] Since we found a unique solution for any given \(z\) we conclude the desired formula.
Consider the inverse stereographic projection \(P^{-1}:\mathbb C\to \mathbb{S}^2\setminus
\left\lbrace N\right\rbrace\).
(a) Show that \(P^{-1}\) takes the
circle \(\left\lbrace z\in\mathbb C\; | \;
{\left|z\right|}=c\right\rbrace\), where \(c> 0\) is a given positive number, to a
circle on \(\mathbb{S}^2\setminus \left\lbrace
N\right\rbrace\) that is parallel to the \(xy-\)plane.
(b)* Explain geometrically why the image of the line \(a \mathop{\text{Re}}\left(
z\right)+b\mathop{\text{Im}}\left( z\right)=0\), where \(a,b\in\mathbb R\) are not both zero, by
\(P^{-1}\) lies on a great circle on
\(\mathbb{S}^2\setminus \left\lbrace
N\right\rbrace\) that passes via the south pole (in fact – it is
the entire circle).
(a) Assuming that \({\left|z\right|}=c\) we find that \[P^{-1}(z) = \left(
\frac{2\mathop{\text{Re}}\left(
z\right)}{1+c^2},\frac{2\mathop{\text{Im}}\left( z\right)}{1+c^2},
\frac{c^2-1}{1+c^2}\right).\] The image of the concentric circle
is on the plane \(z=
\frac{c^2-1}{1+c^2}\), i.e. parallel to the \(xy-\)axis, and since \[\left( \frac{2\mathop{\text{Re}}\left(
z\right)}{1+c^2}\right)^2+\left( \frac{2\mathop{\text{Im}}\left(
z\right)}{1+c^2}\right)^2 = \frac{4{\left|z\right|}^2}{\left(
1+c^2\right)^2} = \frac{4c^2}{\left( 1+c^2\right)^2}\] we
conclude that the image is indeed a circle.
(b) A great circle on \(\mathbb{S}^2\setminus
\left\lbrace N\right\rbrace\) that passes through the south pole
is parametrised by a fixed angle of \(\xi\eta-\)rotation and a free azimuthal
angle. In other words, it is parametrised by \[\begin{aligned}
&\xi(t) = \cos\left( \theta_0\right)\sin \left( t\right),\\
&\eta(t) = \sin\left( \theta_0\right)\sin \left( t\right),\\
&\zeta(t)= \cos \left( t\right),
\end{aligned}\] where \(\theta_0\) is fixed and is determined by
the identity \(\tan\left( \theta_0\right) =
\frac{\eta(t)}{\xi(t)}\) for all \(t\in
(0,\pi]\). The line \(a
\mathop{\text{Re}}\left( z\right)+b\mathop{\text{Im}}\left(
z\right)=0\) satisfies \[\frac{\eta}{\xi} = \frac{\mathop{\text{Im}}\left(
z\right)}{\mathop{\text{Re}}\left( z\right)}=-\frac{a}{b}\] when
\(b\not=0\) and if \(b=0\) we find that \(\mathop{\text{Re}}\left( z\right)=0\) and
consequently \(\xi=0\). We conclude
that the image of the line is indeed on a great circle that passed
through the south pole.
Show that \(P^{-1}(z) = -P^{-1}(w)\) (i.e. the point \(P^{-1}(z)\) and \(-P^{-1}(w)\) are diametrically opposite on the Riemann sphere) if and only if \(w=-\frac{1}{{\bar z}}\).
We have that \(P^{-1}(z) = -P^{-1}(w)\) if and only if \[\frac{2\mathop{\text{Re}}\left( z\right)}{1+{\left|z\right|}^2} = -\frac{2\mathop{\text{Re}}\left( w\right)}{1+{\left|w\right|}^2},\] \[\frac{2\mathop{\text{Im}}\left( z\right)}{1+{\left|z\right|}^2} = -\frac{2\mathop{\text{Im}}\left( w\right)}{1+{\left|w\right|}^2},\] \[\frac{{\left|z\right|}^2-1}{1+{\left|z\right|}^2} = \frac{1-{\left|w\right|}^2}{1+{\left|w\right|}^2}.\] Using the last equation we find that \[\left( {\left|z\right|}^2-1\right)\left( 1+{\left|w\right|}^2\right) = \left( 1-{\left|w\right|}^2\right)\left( 1+{\left|z\right|}^2\right)\] or \[{\left|z\right|}^2+{\left|z\right|}^2{\left|w\right|}^2 -1-{\left|w\right|}^2 = 1-{\left|w\right|}^2+{\left|z\right|}^2-{\left|z\right|}^2{\left|w\right|}^2.\] The above holds if and only if \({\left|z\right|}{\left|w\right|}=1\). Plugging this into the first equation we find that \[\mathop{\text{Re}}\left( w\right) = -\frac{1+{\left|w\right|}^2}{1+{\left|z\right|}^2}\mathop{\text{Re}}\left( z\right) = -\frac{1+\frac{1}{{\left|z\right|}^2}}{1+{\left|z\right|}^2}\mathop{\text{Re}}\left( z\right) = -\frac{\mathop{\text{Re}}\left( z\right)}{{\left|z\right|}^2} = -\mathop{\text{Re}}\left( \frac{z}{{\left|z\right|}^2}\right)=-\mathop{\text{Re}}\left( \frac{1}{{\bar z}}\right).\] Similarly \(\mathop{\text{Im}}\left( w\right) = -\mathop{\text{Im}}\left( \frac{1}{{\bar z}}\right).\) We conclude that \(w=-\frac{1}{{\bar z}}\).