Tutorial: Complex Variables
Francis Wright, June 2018
Click on a problem header to show/hide my REDUCE solution.
- Find all solutions \(z \in \mathbb{C}\) of the equation \(z^3 = 8\).
- Find all solutions \(z \in \mathbb{C}\) of the equation \(e^{3z} = 1\).
- Suppose \(w = f(z) = \frac{z-1}{z-3}\). Show that the image under \(f\) of the line \(Re(z) = 2\) is the unit circle \(\{w \in \mathbb{C} : |w| = 1\}\).
% (a) solve(z^3=8); % (b) solve(exp(3z)=1); % (c) w := (z-1)/(z-3)$ z := 2 + i*y$ w; (w * conj w) where impart y => 0, repart y => y; % since y is real
Find the Taylor series expansion \(\sum_{n=0}^\infty
a_n(z-3)^n\) of \(f(z) = 1/(1+4z)\) about the point \(z_0
= 3\).
clear z; f := 1/(1+4z); % Deduce that the nth derivative is (-4)^n*n!/(1+4z)^(n+1). Hence operator a; let a(~n) => (-4)^n/13^(n+1); on div, revpri; for n := 0:6 sum a(n)*!(z!-3!)^n; % where I have used !(z!-3!) instead of (z-3) to stop the powers being expanded. % Check: load_package "taylor"; taylorprintterms := all$ taylor(f,z,3,6);
If \(g(z) = \frac{1}{(z-1)(3-z)}\), find the Laurent
series \(\sum_{n=0}^\infty a_n(z-1)^n + \sum_{n=1}^\infty
b_n(z-1)^{-n}\) for \(g\) about the point \(z = 1\).
on factor; g := 1/((z-1)*(3-z)); gpf := pf(g,z); load_package "taylor"; taylorprintterms := all$ write "Increasing sum is ", taylor(second gpf, z, 1, 6); write "Decreasing sum is ", first gpf; off factor;
Let \(f(z) = y(3x^2 - y^2) + ix(x^2 - 3y^2)\). Show that
\(f\) is complex differentiable at just one point, and
compute its derivative at this point.
% f(z) = y*(3x^2-y^2) + i*x*(x^2-3y^2) = u(x,y) + i*v(x,y) where z = x * i*y. % Hence u := y*(3x^2-y^2); v := x*(x^2-3y^2); % The Cauchy-Riemann equations are {df(u,x) = df(v,y), df(u,y) = -df(v,x)}; solve ws; sub(ws, df(u + i*v,x));
Locate the poles of the function
\[ f(z) = \frac{z}{(z^2 + 9)^2}. \]
Determine the order of each pole, and its residue.
f := z/(z^2+9)^2; poles := solve(den f); orders := root_multiplicities; z0 := rhs first poles$ ord := first orders$ limit(f*(z-z0)^ord, z, z0); z0 := rhs second poles$ ord := second orders$ limit(f*(z-z0)^ord, z, z0);
Evaluate the integral \(\int_C f(z)\,dz\) when \(f(z) =
\bar{z}\) (the complex conjugate of \(z\)) and
- \(C\) is the straight line segment from \(+2\) to \(-2\).
- \(C\) is the curve from \(+2\) to \(-2\) along the upper half of the radius-\(2\) circle centred at \(0\).
f := conj z$ % (a) let repart x => x, impart x => 0; % since x is real z := x; % x = +2 to x = -2 f; int(f, x, +2, -2); % (b) let repart theta => theta, impart theta => 0; % since theta is real z := 2 exp(i*theta); % theta from 0 to pi f; int(f, theta, 0, pi);
Use Cauchy's Theorem to calculate
\[ \int_C \frac{\sin(z)}{z^2 + 2z + 2} \, dz \]
where \(C\) is the positively oriented circle centred at the origin
with radius \(1\), being careful to justify your answer.
clear z; on complex; operator f; let f(~z) => sin(z)/(z^2+2z+2); poles := solve(den f(z)); orders := root_multiplicities; z0 := rhs first poles; abs z0; % > 1 so outside C z0 := rhs second poles; abs z0; % > 1 so outside C % No poles inside (or on) C, so integral = 0.