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Section 7.6 Sample exercises

Show that \(\Re(z) = \frac{z+\bar{z}}{2}\) and \(\Im(z) = \frac{z-\bar{z}}{2i}\text{.}\)

Solution.

If \(z=x+iy\text{,}\) then

\begin{equation*} \frac{z+\bar{z}}{2} = \frac{z+iy+ z-iy}{2} = \frac{2x}{2} = x=\Re(z) \end{equation*}

and

\begin{equation*} \frac{z-\bar{z}}{2i} = \frac{z+iy-( z-iy)}{2i} = \frac{2iy}{2i} = y=\Im(z)\text{.} \end{equation*}

Show that

\begin{equation*} \Re(zw)\leq \frac{|z|^2+|w|^2}{2}\text{.} \end{equation*}
Solution.

By Lemma 7.1.4, and the AM-GM inequality,

\begin{equation*} \Re(zw) \leq |zw| =|z|\cdot |w| =(|z|^2\cdot |w|^2)^{\frac{1}{2}}\leq \frac{|z|^2+|w|^2}{2}\text{.} \end{equation*}

Solve \(z^2=i\overline{z}\text{.}\)

Solution.

Notice that if this equation holds, then

\begin{equation*} |z|^2=|i\overline{z}|=|i|\cdot |\overline{z}| = |z|\text{,} \end{equation*}

so either \(|z|=0\) or \(|z|=1\text{.}\) In the latter case, this means that \(\overline{z} = \frac{1}{z}\text{,}\) and so

\begin{equation*} z^2=i\overline{z} = \frac{i}{z} \end{equation*}

which implies

\begin{equation*} z^3=i\text{.} \end{equation*}

Hence, we just need to solve this equation now. Since \(i=e^{\frac{\pi}{2}i}\text{,}\) then \(e^{\frac{\pi}{6}i}\) is one solution. Thus, to get all 3 solutions, we multiply this by all the 3rd roots of unity, so we get

\begin{equation*} e^{\frac{\pi}{6}i}, \;\; e^{\frac{5\pi}{6}i}, \;\; e^{\frac{3\pi}{2}i}\text{.} \end{equation*}

Thus, all solutions to the original equation are

\begin{equation*} 0, \;\; \pm \frac{\sqrt{3}}{2}+\frac{i}{2}, \;\; -i\text{.} \end{equation*}

Solve \(|z|^2 - z|z| + z = 0\text{.}\)

Solution.

Note that by rearranging the equations so that the \(z's\) are on one side and \(|z|'s\) are on the other, we get if \(|z|\neq 1\) that

\begin{equation*} z =\frac{|z|^{2}}{|z|-1}\text{,} \end{equation*}

and so \(z\in\R\text{.}\) If \(z\geq 0\text{,}\) then our original equation becomes \(z^2-z^2+z=0\text{,}\) hence \(z=0\text{.}\) If \(z\lt 0\text{,}\) then \(|z|=-z\text{,}\) and our original equation is

\begin{equation*} z^2+z^2+z=0 \end{equation*}

So \(0=z(2z+1)\text{,}\) hence \(z=0\) or \(z=-\frac{1}{2}\text{.}\)

If \(|z|=1\text{,}\) then the original equation is

\begin{equation*} 1-z+z=0\text{,} \end{equation*}

which is impossible, so there are no solutions in this case. Thus, \(z=0,-\frac{1}{2}\) are the only solutions.

Show that if \(|z| = 1\text{,}\) then \(\Re\frac{1}{1-z} = \frac{1}{2}\text{.}\)

Solution.
\begin{equation*} \frac{1}{1-z} = \frac{1}{1-z}\frac{1+\bar{z}}{1+\bar{z}} = \frac{1+\bar{z}}{1-z+\bar{z} + z\bar{z}} \end{equation*}
\begin{equation*} = \frac{1+\bar{z}}{1-2iy-|z|^{2}}=\frac{1+\bar{z}}{-2iy}=i\frac{1+\bar{z}}{2y} \end{equation*}
\begin{equation*} \Re \frac{1}{1-z} = \Re i\frac{1+\bar{z}}{2y} = \Re i\frac{1+x-iy}{2y} = \Re\left(\frac{y}{2y} + i\frac{1+x}{2y}\right) = \frac{y}{2y} \end{equation*}
\begin{equation*} =\frac{1}{2} \end{equation*}

Describe geometrically the points \(z\in\mathbb{C}\) so that \(|z-1|=|z+i|\text{.}\)

Solution.

These are all points that lie on the \(x=y\) line in the \(xy\)-plane (so the line that makes a 45 degree angle with the axes. To see this, let \(z=x+i\) and observe that

\begin{align*} |z-1|=|z+i| \amp \;\;\; \Longleftrightarrow \;\;\; |z-1|^2=|z+i|^2\\ \amp \;\;\; \Longleftrightarrow \;\;\; (z-1)\overline{(z-1)} = (z+i)\overline{(z+i)}\\ \amp \;\;\; \Longleftrightarrow \;\;\; (z-1)(\bar{z}-1)=(z+i)(\bar{z}-i)\\ \amp \;\;\; \Longleftrightarrow \;\;\; z\bar{z}-z-\bar{z}+1 = z\bar{z} -zi+i\bar{z}+1\\ \amp \;\;\; \Longleftrightarrow \;\;\; -z-\bar{z}= -zi+i\bar{z}\\ \amp \;\;\; \Longleftrightarrow \;\;\; -(x+iy)-(x-iy) = -(x+iy)i+i(x-iy)\\ \amp \;\;\; \Longleftrightarrow \;\;\; -2x = -2y\\ \amp \;\;\; \Longleftrightarrow \;\;\; x=y\text{.} \end{align*}

Find all solutions to \((z+1)^4=z^4\text{.}\)

Solution.

First, we write this equation as \((1+1/z)^4=1\text{,}\) so this implies that \(1+1/z\) is one of the \(4\)th roots of unity \(\pm1,\pm i\text{.}\) However, there are no solutions to \(1+1/z=1\text{,}\) so the only solutions are when \(1+1/z\) is \(-1\) or \(\pm i\text{,}\) in which case

\begin{equation*} z=-\frac{1}{2}, \frac{1}{-1\pm i}\text{.} \end{equation*}

Let \(\mathbb{D}=\{z\in \mathbb{C}: |z|\lt 1\}\text{,}\) that is, the set of complex numbers with modulus strictly less than \(1\text{.}\) \(\mathbb{D}\) is for “disc”. For a complex number \(z\in \mathbb{D}\) and a real number \(-1\lt r\lt 1\text{,}\) define 1 

\begin{equation*} f_{r}(z) = \frac{z-r}{1-zr}\text{.} \end{equation*}
The function \(f_{r}\) is called a Möbius transformation and is usually defined for \(r\in \mathbb{D}\) as well, not just \(-1\lt r\lt 1\text{.}\) They are the only functions with the property that, if \(A\) is a circle or straight line in \(\mathbb{D}\text{,}\) then \(f_{r}(A)\) is also a line or circle (this is not part of the problem, it's just cool).

Show \(f_{r}(z)\in \mathbb{D}\) for all \(z\in \mathbb{D}\) and \(-1\lt r\lt 1\text{.}\)

Solution.

If \(-1\lt r\lt 1\text{,}\) then

\begin{align*} \left|\frac{z-r}{1-rz}\right|\lt 1 \amp \Leftrightarrow \;\;|z-r|\lt |1-rz|\\ \amp \Leftrightarrow \;\; |z-r|^{2}\lt |1-rz|^{2}\\ \amp \Leftrightarrow \;\; |z|^{2}+|r|^{2}-2Re(zr)\lt 1+|rz|^{2}-2Re(zr)\\ \amp \Leftrightarrow \;\; |z|^{2}+|r|^{2}-|rz|^{2}\lt 1\\ \amp \Leftrightarrow \;\; |z|^{2}(1-|r|^{2})+|r|^{2}\lt 1 \end{align*}

But this inequality is true since \(|z|\lt 1\text{,}\) so

\begin{equation*} |z|^{2}(1-|r|^{2})+|r|^{2}\lt 1\cdot (1-|r|^{2})+|r|^{2}=1\text{.} \end{equation*}