MTH2003 Complex Analysis Practical 2

• Practical Sheet.

Question 1

…

Question 2

Given the function with real and imaginary parts…

$$u(x,y)=\sin x\cosh y\text{and}v(x,y)=\cos x\sinh y$$

And the Cauchy-Riemann equations…

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\text{and}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

We can test the function satisfies these C-R equations by calculating…

$$\begin{array}{rl}\frac{\partial u}{\partial x}& =\cos x\cosh y\\ \frac{\partial v}{\partial y}& =\cos x\cosh y\therefore \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\square \\ \frac{\partial u}{\partial y}& =\sin x\sinh y\\ -\frac{\partial v}{\partial x}& =-\sin x\sinh y\therefore \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\square \end{array}$$

Question 3

Given the following functions of $u(x,y)$, we can find their harmonic conjugates $v(x,y)$ (using the C-R equations) and analytic functions (by summing the conjugates) $f(z)$ of which $u$ and $v$ are the real and imaginary parts…

Part A

$$u(x,y)=e^{-x}\cos y:\frac{\partial u}{\partial x}=-e^{-x}\cos y=\frac{\partial v}{\partial y}⟹\boxed{v(x,y)=-e^{-x}\sin y}$$ $$\boxed{\therefore e^{-x}\cos y-e^{-x}\sin y}$$

note: should also include the function of integration!! (this is different to normal calculus, as it’s multivariable)

also note that the final function should be expressed in terms of $z$!!

Part B

$$u(x,y)=y^2-x^2:\frac{\partial u}{\partial x}=y^2-2x=\frac{\partial v}{\partial y}⟹\boxed{v(x,y)=\frac{1}{3}{y}^3-2x}$$ $$\boxed{\therefore y^2-x^2+\frac{1}{3}{y}^3-2x}$$

note: should also include the function of integration!! (this is different to normal calculus, as it’s multivariable)

also note that the final function should be expressed in terms of $z$!!

Part C

$$u(x,y)=\frac{y}{x^2+y^2}:\frac{\partial u}{\partial x}=...=\frac{\partial v}{\partial y}⟹\boxed{v(x,y)=...}$$

Left for future revision

Note: include this idea of harmonic conjugates in the lecture notes.

Question 4

Given that we have the real and imaginary parts of two functions which satisfy the C-R equations on specified regions, we can state their analytic function form…

$$\begin{array}{rl}u(x,y)& =x^3-3x{y}^2\\ v(x,y)& =3x^{2}y-y^3\end{array}\text{for all }z\in \mathbb{C}⟹f(x,y)=x^3-3x{y}^2+i(3x^{2}y-y^3)$$

Defined on the union of its constituent intervals (for all $z\in \mathbb{C}$).

$$\begin{array}{rl}u(x,y)& =\frac{x}{x^2+y^2}\text{for all }z\in \mathbb{C}\\ v(x,y)& =-\frac{y}{x^2+y^2}\text{for all }{x}^2+y^2\ne 0\end{array}⟹f(x,y)=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}$$

Defined on the union of its constituent intervals (for all $x^2+y^2\ne 0$).

Question 5

Using the Cauchy-Riemann equations, we can show that the following are not differentiable anywhere:

$$f(z)=\text{Im}(z)=y:\frac{\partial u}{\partial x}=0\ne 1=\frac{\partial v}{\partial y}$$ $$f(z)=|z|=|x+iy|=\sqrt{x^2+y^2}:\frac{\partial u}{\partial x}=...\ne 0=\frac{\partial v}{\partial y}$$ $$f(z)=\text{arg}(z):...$$

Left for future revision

Question 6

Given the following functions, we can find values for $z$ which allow them to satisfy the C-R equations…

$$w=z^2=(x+iy{)}^2=(x^2-y^2)+i(2xy):\frac{\partial u}{\partial x}=2x-y^2=2x=\frac{\partial v}{\partial y}⟹\dots$$ $$w=|z{|}^2=(x^2+y^2)+i(0)$$

Left for future revision.

Question 7

Given a function defined as $f(z)=\ln r+i\theta$ for $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$, we can first convert it to Cartesian coordinates:

$$r=\sqrt{x^2+y^2},\theta =\arctan \left(\frac{y}{x}\right)$$

etc.?

And then show how it satisfies the C-R equations for all $z$ such that $\text{Re}(z)>0$.

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