Analytic Functions & Cauchy-Riemann Equations

Solution:

Write $f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$. Then $u(x,y) = x^2 - y^2$ and $v(x,y) = 2xy$.

Compute partial derivatives: $u_x = 2x$, $u_y = -2y$, $v_x = 2y$, $v_y = 2x$.

Check CR equations: $u_x = 2x = v_y$ ✓ and $u_y = -2y = -v_x$ ✓

Therefore, $f'(z) = u_x + iv_x = 2x + 2iy = 2z$, as expected.

Solution:

Check: $u_{xx} = 2$, $u_{yy} = -2$, so $u_{xx} + u_{yy} = 0$ ✓

To find harmonic conjugate $v$, use CR equations: $u_x = 2x = v_y$ and $u_y = -2y = -v_x$

From $v_y = 2x$: $v = 2xy + g(x)$ for some function $g$

From $v_x = 2y$: $2y + g'(x) = 2y$, so $g'(x) = 0$, hence $g(x) = C$

Therefore $v(x,y) = 2xy + C$. Taking $C = 0$, we get $f(z) = z^2$.

Solution:

Write $f(z) = |z|^2 = x^2 + y^2$, so $u = x^2 + y^2$, $v = 0$

Then $u_x = 2x$, $u_y = 2y$, $v_x = 0$, $v_y = 0$

CR equations require: $2x = 0$ and $2y = 0$, so only at $z = 0$

Check: $f'(0) = \lim_{z \to 0} \frac{|z|^2}{z} = \lim_{z \to 0} \bar{z} = 0$

Therefore, $f$ is differentiable only at $z = 0$, and nowhere analytic.

Solution:

Write $f(z) = \overline{z}^2 = (x - iy)^2 = x^2 - y^2 - 2ixy$

So $u = x^2 - y^2$ and $v = -2xy$

$u_x = 2x$, $u_y = -2y$, $v_x = -2y$, $v_y = -2x$

CR equations: $u_x = 2x \neq v_y = -2x$ (unless $x = 0$)

$u_y = -2y \neq -v_x = 2y$ (unless $y = 0$)

CR equations only hold at $z = 0$, so $f$ is not analytic anywhere.

Solution:

Approach along $\Delta z = re^{i\theta}$ for fixed $\theta$:

Using chain rule: $= e^{-i\theta}[u_x \cos\theta + u_y \sin\theta + i(v_x \cos\theta + v_y \sin\theta)]$

This must equal $u_x + iv_x$ (the derivative along real axis) for all $\theta$. Setting $\theta = 0$ and $\theta = \pi/2$ gives the CR equations.

Solution:

First verify $u$ is harmonic: $u_x = \frac{2x}{x^2 + y^2}$, $u_{xx} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2}$

$u_y = \frac{2y}{x^2 + y^2}$, $u_{yy} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}$

$u_{xx} + u_{yy} = 0$ ✓

Using CR equations: $v_y = u_x = \frac{2x}{x^2 + y^2}$, so $v = 2\arctan(y/x) + g(x) = 2\arg(z) + g(x)$

$-v_x = u_y = \frac{2y}{x^2 + y^2}$ gives $g'(x) = 0$, so $v(x,y) = 2\arg(z) + C$

Taking $C = 0$, we get $f(z) = \ln|z| + 2i\arg(z) = 2\ln z$ (in a suitable branch).

Solution:

$u_{xx} = e^x\cos y$, $u_{yy} = -e^x\cos y$, so $u_{xx} + u_{yy} = 0$ ✓

Using CR: $v_y = u_x = e^x\cos y$, so $v = e^x\sin y + g(x)$

$v_x = e^x\sin y + g'(x) = -u_y = e^x\sin y$, so $g'(x) = 0$, $g(x) = C$

$v(x,y) = e^x\sin y$ (taking $C = 0$)

Therefore $f(z) = e^x\cos y + ie^x\sin y = e^{x+iy} = e^z$

Solution:

In polar form: if $z = re^{i\theta}$, then $f(z) = r^2 e^{i2\theta}$

The function squares the modulus and doubles the argument. This means:

• Circles centered at origin map to circles (radius squared)
• Rays from origin map to rays (angle doubled)
• Angles are preserved, but at $z = 0$ where $f'(0) = 0$, angles are doubled

Solution:

If $u$ is constant, then $u_x = u_y = 0$

By CR equations: $v_y = u_x = 0$ and $v_x = -u_y = 0$

Therefore $v$ is also constant, so $f = u + iv$ is constant.

Solution:

$f(z) = z + \overline{z} = (x + iy) + (x - iy) = 2x$

So $u = 2x$, $v = 0$

$u_x = 2$, $u_y = 0$, $v_x = 0$, $v_y = 0$

CR requires $u_x = 2 = v_y = 0$, which is impossible.

Therefore $f$ is nowhere analytic.

Solution:

Write $f(z) = \frac{1}{z} = \frac{\overline{z}}{|z|^2} = \frac{x - iy}{x^2 + y^2} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}$

So $u = \frac{x}{x^2 + y^2}$, $v = -\frac{y}{x^2 + y^2}$

Computing derivatives:

$u_x = \frac{y^2 - x^2}{(x^2 + y^2)^2}$, $u_y = \frac{-2xy}{(x^2 + y^2)^2}$

$v_x = \frac{2xy}{(x^2 + y^2)^2}$, $v_y = \frac{x^2 - y^2}{(x^2 + y^2)^2}$

Check CR: $u_x = \frac{y^2 - x^2}{(x^2 + y^2)^2} = v_y$ ✓

$u_y = \frac{-2xy}{(x^2 + y^2)^2} = -v_x$ ✓

CR equations are satisfied for $z \neq 0$, so $f$ is analytic there.

Solution:

Write $f(z) = \frac{z}{\overline{z}} = \frac{z^2}{|z|^2} = \frac{(x+iy)^2}{x^2+y^2} = \frac{x^2-y^2+2ixy}{x^2+y^2}$

So $u = \frac{x^2-y^2}{x^2+y^2}$, $v = \frac{2xy}{x^2+y^2}$ (for $z \neq 0$)

Computing partial derivatives and checking CR equations shows they are not satisfied anywhere except possibly at isolated points. Therefore, $f$ is nowhere differentiable.

Solution:

If $u^2 + v^2 = C$ (constant), differentiate with respect to $x$:

$2uu_x + 2vv_x = 0$, so $uu_x + vv_x = 0$

Using CR equations: $uu_x - vu_y = 0$ (since $v_x = -u_y$)

Similarly, $uu_y + vv_y = 0$ gives $uu_y + vu_x = 0$

Solving these, if $u^2 + v^2 \neq 0$, we get $u_x = u_y = 0$, so $u$ is constant, and by CR equations, $v$ is also constant.

Solution:

If $f = u + iv$, then $\overline{f(\overline{z})} = \overline{u(x,-y) + iv(x,-y)} = u(x,-y) - iv(x,-y)$

The condition implies $u(x,y) = u(x,-y)$ and $v(x,y) = -v(x,-y)$.

If $f$ is analytic, its real and imaginary parts are harmonic and satisfy these symmetry conditions. Functions of the form $f(z) = g(z) + \overline{g(z)}$ where $g$ is analytic satisfy this.

Solution:

If $|f|^2 = u^2 + v^2 = C$ is constant, then by the previous example, $f$ is constant.

Alternatively, if $|f| = C$ and $f \neq 0$, then $1/f$ is analytic and $|1/f| = 1/C$. By maximum modulus principle, both $|f|$ and $|1/f|$ are constant, so $f$ must be constant.

Solution:

Write $z^n = r^n e^{in\theta} = r^n\cos(n\theta) + ir^n\sin(n\theta)$

So $u = r^n\cos(n\theta)$, $v = r^n\sin(n\theta)$

$\frac{\partial u}{\partial r} = nr^{n-1}\cos(n\theta) = \frac{1}{r} \cdot nr^n\sin(n\theta) = \frac{1}{r}\frac{\partial v}{\partial\theta}$ ✓

$\frac{\partial v}{\partial r} = nr^{n-1}\sin(n\theta) = -\frac{1}{r} \cdot (-nr^n\cos(n\theta)) = -\frac{1}{r}\frac{\partial u}{\partial\theta}$ ✓

Solution:

In polar form: $\ln z = \ln r + i\theta$ where $z = re^{i\theta}$

So $u(r,\theta) = \ln r$, $v(r,\theta) = \theta$

$u_r = 1/r$, $v_\theta = 1$, so $u_r = v_\theta/r$ ✓

$u_\theta = 0$, $v_r = 0$, so $v_r = -u_\theta/r$ ✓

Solution:

From CR equations: $u_x = v_y$ and $u_y = -v_x$

Differentiating: $v_{xx} = -u_{yx} = -u_{xy} = -(v_{yy})$ (using equality of mixed partials)

Therefore: $v_{xx} + v_{yy} = 0$, so $v$ is harmonic.

Solution:

First verify $u$ is harmonic: $u_{xx} = 6x$, $u_{yy} = -6x$, so $u_{xx} + u_{yy} = 0$ ✓

From CR equations: $v_y = u_x = 3x^2 - 3y^2$

Integrating with respect to $y$: $v = 3x^2y - y^3 + g(x)$ for some function $g(x)$

Also: $v_x = -u_y = 6xy$

Differentiating our expression: $v_x = 6xy + g'(x) = 6xy$

So $g'(x) = 0$, hence $g(x) = C$ (constant)

Therefore: $v(x,y) = 3x^2y - y^3 + C$

The analytic function is: $f(z) = u + iv = x^3 - 3xy^2 + i(3x^2y - y^3) + iC = z^3 + iC$

Solution:

Here $u = e^x\cos y$, $v = e^x\sin y$

Check CR equations:

$u_x = e^x\cos y$, $v_y = e^x\cos y$ → $u_x = v_y$ ✓

$u_y = -e^x\sin y$, $v_x = e^x\sin y$ → $u_y = -v_x$ ✓

Since partial derivatives are continuous everywhere, $f$ is analytic everywhere.

The derivative: $f'(z) = u_x + iv_x = e^x\cos y + ie^x\sin y = e^x(\cos y + i\sin y) = f(z)$

Note: This is $f(z) = e^z$ in disguise, confirming that $(e^z)' = e^z$.

Solution:

Write $f(z) = \frac{1}{z} = \frac{\overline{z}}{|z|^2} = \frac{x - iy}{x^2 + y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$

So $u = \frac{x}{x^2+y^2}$, $v = -\frac{y}{x^2+y^2}$

Computing partial derivatives:

$u_x = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$

$v_y = -\frac{(x^2+y^2) - y(2y)}{(x^2+y^2)^2} = -\frac{x^2 - y^2}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$

So $u_x = v_y$ ✓

$u_y = \frac{-x(2y)}{(x^2+y^2)^2} = \frac{-2xy}{(x^2+y^2)^2}$

$v_x = \frac{y(2x)}{(x^2+y^2)^2} = \frac{2xy}{(x^2+y^2)^2}$

So $u_y = -v_x$ ✓

Since partial derivatives are continuous for $z \neq 0$, $f$ is analytic in $\mathbb{C} \setminus \{0\}$.

Solution:

Given $u(x,y) = x^2 - y^2 + 2x$

Verify harmonic: $u_{xx} = 2$, $u_{yy} = -2$, so $u_{xx} + u_{yy} = 0$ ✓

From CR: $v_y = u_x = 2x + 2$

Integrating: $v = 2xy + 2y + g(x)$

Also: $v_x = -u_y = 2y$

Differentiating: $v_x = 2y + g'(x) = 2y$, so $g'(x) = 0$, hence $g(x) = C$

Therefore: $f(z) = x^2 - y^2 + 2x + i(2xy + 2y) + iC$

Simplifying: $f(z) = (x^2 - y^2 + 2ixy) + 2(x + iy) + iC = z^2 + 2z + iC$

Using $f(0) = 0$: $0 + 0 + iC = 0$, so $C = 0$

Final answer: $f(z) = z^2 + 2z$