Practice midterm 1 solutions

As usual, be sure to include your method, and remember to write your name.

Problem 1. Find a square root of $1-i$.

The above problem is directed towards Objective 1 (complex numbers, algebra, and functions).

Solution. We can write $1-i=\sqrt{2}\cdot e^{-\pi i/4}$, so $\sqrt{1-i}=\sqrt[4]{2}\cdot e^{-\pi i/8}$. Another possibility would be $-\sqrt[4]{2}\cdot e^{-\pi i/8}=\sqrt[4]{2}\cdot e^{7\pi i/8}$.

Problem 2. Where are the branch points of $\frac{\sqrt{z^2-4}}{z}$? What are the phase factors at each?

The above problem is directed towards Objective 1 (complex numbers, algebra, and functions).

Solution. There are three natural points to worry about: the two points $z_0=\pm 2$ where $z_0^2-4=0$, and the point $z_0=0$ where the function is undefined. Near $z_0=2$, we can write the function as $\frac{\sqrt{z+2}}{z}\cdot \sqrt{z-2}$ and the first factor is continuous near $z_0=2$, so we only have to worry about the second factor, which has phase factor $e^{2\pi i\cdot \frac{1}{2}}=-1$. Similarly at $z_0=-2$ we again get phase factor $-1$. At $z_0=0$, we can write the function as $\sqrt{z^2-4}\cdot z^{-1}$, with the first factor continuous near $z_0=0$; we know that the second factor has phase factor $e^{2\pi i\cdot -1}=1$, so $z_0=0$ is actually not a branch point, and we just have $z_0=\pm 2$, both with phase factor $-1$.

Problem 3. Show that if $f:ℂ\to ℂ$ is an analytic function such that $\mathrm{Re}(f(z))=\mathrm{Im}(f(z))$ for all $z$, then $f$ is constant.

The above problem is directed towards Objective 2 (analytic functions and the Cauchy–Riemann equations).

Solution. If $f=u+iv$, then the condition is $u=v$. Since $f$ is analytic, it satisfies the Cauchy–Riemann equations, so $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial x}$ since $v=u$, and then by the Cauchy–Riemann equations again this is further equal to $-\frac{\partial v}{\partial y}=-\frac{\partial u}{\partial y}$. Therefore $\frac{\partial u}{\partial y}=-\frac{\partial u}{\partial y}$, so $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=0$. Therefore $u$ is constant, and since $v=u$ it must also be constant, so $f$ is constant.

Problem 4. Verify using the Cauchy–Riemann equations that $f(z)=z^2$ is analytic on $ℂ$.

The above problem is directed towards Objective 2 (analytic functions and the Cauchy–Riemann equations).

Solution. If $z=x+iy$, then $z^2=x^2+2ixy-y^2$, so if $f=u+iv$ then $u(x,y)=x^2-y^2$ while $v(x,y)=2xy$. Therefore ${\partial }_{x}u=2x$, ${\partial }_{y}u=-2y$, ${\partial }_{x}v=2y$, and ${\partial }_{y}v=2x$. Therefore we have ${\partial }_{x}u={\partial }_{y}v$ and ${\partial }_{y}u=-{\partial }_{x}v$.

Problem 5. Let $h(z)={|z|}^2$, as a function $ℂ\to ℂ$. Show that $h$ does not satisfy the mean value property.

The above problem is directed towards Objective 3 (harmonic functions).

Solution. Although one can verify by explicit integration, the easiest thing is to observe that $h(x+iy)=x^2+y^2$ is not harmonic, since $\mathrm{\Delta }h=4\ne 0$, and so cannot satisfy the mean value property on $ℂ$, since if it did it would be harmonic.

Problem 6. Let , and let $u:𝔻\to ℝ$ be given by $u(x,y)=x^2-y^2$. Show that $u$ is harmonic, and determine whether or not it has a harmonic conjugate; if so, find one.

The above problem is directed towards Objective 3 (harmonic functions).

Solution. First, $\mathrm{\Delta }u=\frac{{\partial }^2}{\partial x^2}(x^2-y^2)+\frac{{\partial }^2}{\partial y^2}(x^2-y^2)=2-2=0$, so $u$ is harmonic. Second, since $𝔻$ is star-shaped, $u$ must have a harmonic conjugate $v$. To find it, we observe that $v$ must satisfy $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=2y$ and $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2x$, so $v(x,y)=2xy+C$ for some constant $C$. (Just writing e.g. $v(x,y)=2xy$ would also be acceptable, since the problem only asks for a harmonic conjugate.)