Basic Tricks

• For polynomials with real coefficients, every complex solution \(\lambda\) has a twin solution \(\lambda^\star\)
• Using the conjugate to quickly calculate \(\textrm{Re}(z)\) and \(\textrm{Im}(z)\) can be done as the following:

\[\textrm{Re}(z)=\frac{z+z^\star}{2}\qquad\&\qquad\textrm{Im}(z)=\frac{z-z^\star}{2i}\]

• Another trick, is to calculate the reciprocal of a complex number:

\[\frac{1}{z}=\frac{z^\star}{|z|^2}\]

Circuits

If we have an AC circuit with capacitors or inductors in it, then we can analyse the circuit using complex numbers.

• Inductor: stores energy in a magnetic field in an AC circuit; opposes changes in current.
• Capacitor: stores electrical energy in an AC circuit; opposes changes in voltage

In this case we calculate the total impedance, \(Z_{Tot}\). The impedance depends on the angular frequency of the alternating current: \(\omega=2\pi f\), and the components:

• For a resistor, the impedance \(Z=R\), so the contribution to \(Z_{Tot}\) is unchanged.
• For an inductor, with inductance \(L\) and at an angular frequency \(\omega\) then the inductor has impedance \(Z=iL\omega\).
• For a capacitor, with capacitance \(C\) and at an angular frequency \(\omega\) then the capacitor has impedance \(Z=\frac{1}{i\omega C}=\frac{-i}{\omega C}\).

Complex Number Forms

\[ z=x+iy=r(\cos\theta+i\sin\theta)=re^{i\theta} \] Convert polar to Cartesian with: \[ x=r\cos\theta\qquad\&\qquad y=r\sin\theta \] Convert Cartesian to polar with: \[ r=\sqrt{x^2+y^2}=|z|\qquad\&\qquad\tan\theta=y/x \] (adding or subtracting \(\pi\) to get into correct quadrant)

Sine and Cosine

We define \(\sin(x)\) and \(\cos(x)\) as the following, using the consequences of Euler’s Formula. \[ \cos(x)=\frac{e^{ix}+e^{-ix}}{2}\qquad\&\qquad\sin(x)=\frac{e^{ix}-e^{-ix}}{2i} \]

Arc functions

We can use find the formulas for the arc functions, by using their normal respective functions and a clever use of quadratics. \[\begin{align} &\textrm{Let}\;\;x=\cos(y)=\frac{e^{iy}+e^{-iy}}{2}\\ &2x=e^{iy}+e^{-iy}\\ &e^{iy}-2x+e^{-iy}=0\\ &\left(e^{iy}\right)^2-2xe^{iy}+1=0\\ &\textrm{Quadratic Equation}\\ &e^{iy}=x\pm\sqrt{x^2-1}\\ &\therefore iy=\ln(x\pm\sqrt{x^2-1})\\ &\implies y=-i\ln(x\pm\sqrt{x^2-1})=\arccos(x) \end{align}\]

De Moivre’s Theorem

We know that \(\left(e^{i\theta}\right)^n=e^{in\theta}\) for \(n\in\mathbb{Z}\) and \(n\geq 0\), hence: \[ (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta) \] This is De Moivre’s Theorem and is used to aid trigonometric algebra. This can be demonstrated without the use of Euler’s theorem. (It predates Euler’s theorem) You can work out triple angle formulas etc using this method: \[ \cos(3\theta)+i\sin(3\theta)=(\cos\theta+i\sin\theta)^3 \] From real and imaginary parts of the expansion: \[\begin{align} &\cos(3\theta)=\cos^3\theta-3\cos\theta\sin^2\theta\\ &\sin(3\theta)=3\cos^2\theta\sin\theta-\sin^3\theta \end{align}\]

Binomial Expansion

For large values of \(n\) expanding brackets can be tedious. However we can use the binomial expansion: \[ (a+b)^n=\sum^n_{r=0}C^n_ra^{n-r}b^r \] ...Can use \(nCr\) on the calculator, or \[ C^n_r=\frac{n!}{r!(n-r)!} \] Remember that \(0!=1\)

Roots of Unity

Example: Find the roots of \(z^3=1\) \[\begin{align} &\textrm{Obviously, we have}\;z=1\\ \\ &z^3=e^{i3\theta}=\cos(3\theta)+i\sin(3\theta)=1\\ \\ &\textrm{Re:}\;\cos(3\theta)=1\\ &3\theta=2n\pi\\ &\theta=\frac{2n\pi}{3}\\ \\ &\implies z=e^{2\pi i/3},e^{-2\pi i/3}, e^0(=1) \end{align}\]

More Generally

Since \(e^{2\pi ki}=1\) for \(k\in\mathbb{Z}\), then: \[ 1^{\frac{1}{n}}=e^{\frac{2\pi ki}{n}} \] There are an infinite number of these (one for each \(k\)), but only \(n\) of them are different. These are the roots of: \[ z^n-1=0 \]

Does \((a^b)^c=a^{bc}\)?

No! \[\begin{align} &\ln(a^b)=b\ln a\\ &a^b=\exp(b\ln a+2\pi inb)\;\textrm{for}\;n\in\mathbb{Z}\\ \\&\textrm{Hence:}\\ &(a^b)^c=\left\{\exp(b\ln a+2\pi inb)\;| \;n\in\mathbb{Z}\right\}^c\\ &(a^b)^c=\left\{\exp(cb\ln a+2\pi inb+2\pi imc)\;| \;n,m\in\mathbb{Z}\right\}\\ \end{align}\]

Complex Differentiation

\(i\) in functions is just treat like a variable when it comes to differentiation, and \(i^2=-1\). Example Differentiate \(e^{i\alpha t}\) with respect to \(t\): \[ \frac{d}{dt}e^{i\alpha t}=i\alpha e^{i\alpha t} \]

Complex Integration

Its the same as above. Example: Evaluate: \(\int^4_0(t^2+i\sqrt{t})dt\) \[ \int^4_0(t^2+i\sqrt{t})dt=\left[\frac{t^3}{3}+\frac{2it^{3/2}}{3}\right]^4_0=\frac{64}{3}+\frac{16}{3}i \]

Complex ODE's

An ODE can be a complex differential equation for \(z(t)\), for example: \[ a(t)\ddot{z}+b(t)\dot{z}+c(t)z=d(t) \] In this case, we write \(z:\mathbb{R}\to\mathbb{C}\). That is \(z\) takes a real number \(t\) and gives a complex number \(z(t)\). The equation above is equivalent to two real differential equations: \[ \Re(LHS)=\Re(RHS)\qquad\&\qquad\Im(LHS)=\Im(RHS) \] Example: Find the real and imaginary parts of: \[m\ddot{z}+2mf\dot{z}+kz=F_0e^{i\alpha t}\] We can use the definition of a complex number (\(z=x+iy\)): \[ m(\ddot{x}+i\ddot{y})+2mf(\dot{x}+i\dot{y})+k(x+iy)=F_0(\cos(\alpha t)+i\sin(\alpha t)) \] So we therefore get two ODE's \[ \Re:\quad m\ddot{x}+2mf\dot{x}+kx=F_0\cos(\alpha t) \] and... \[ \Im:\quad m\ddot{y}+2mf\dot{y}+ky=F_0\sin(\alpha t) \] To solve these, see Phys113-Series and Differential Equations.

Hyperbolic Trig Functions: Cosh and Sinh

The easy way to remember this is to just take the exponential definitions of \(\sin\) and \(\cos\), remove all \(i\)'s from them, then add on a h to their name! \[ \cosh\theta=\frac{e^\theta+e^{-\theta}}{2}\qquad\&\qquad\sinh\theta=\frac{e^\theta-e^{-\theta}}{2} \] They are the odd and even parts (respectively) of \(e^\theta\)

Important Identity

\[ \cosh^2\theta-\sinh^2\theta=1 \]

Taylor Series

You can use the Taylor series of \(e^x\) and \(e^{-x}\), to work out the hyperbolics equivalent. \[ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \] \[ e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots \] \[ \cosh\theta=\frac{e^\theta+e^{-\theta}}{2}\qquad\&\qquad\sinh\theta=\frac{e^\theta-e^{-\theta}}{2} \] Hence \[ \cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\ldots \] \[ \sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\ldots \]

Comparing Hyperbolic and Circular Trig. Functions

Cauchy-Riemann Equations For Complex Functions

A function is analytic if it can be infinitely differentiated at every point. An analytic function also has a Taylor Series that converges to the function within the radius of convergence around some point \(x_0\). Let \(f(x,y)=u(x,y)+iv(x,y)\) and let \(z=x+iy\), then we have a pair of real functions \(u(x,y)\) and \(v(x,y)\). These can each be differentiated to give the four partial derivatives \[ \frac{\partial u}{\partial x}, \frac{\partial v}{\partial y},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x} \] If \(f\) is analytic then these four quantities satisfy the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\qquad\textrm{and}\qquad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \] All analytic functions have the following properties:

• They are infinitely differentiable.
• They have Taylor’s expansion about each point.
• The radius of convergence of the Taylor’s expansion is the distance to the nearest singularity (place were the function beaks down) in the complex plane.
• For example the function \(f(z) =1+z^2\) has a Taylor series \(f(z)=\sum^\infty_{r=0}(−z^2)^r\). The radius of convergence for the series is \(1\), which corresponds to the distance between the origin and \(i\) (or \(−i\)) which are the places where the function \(f(z)\) breaks down.

Thus knowing how an analytic function behaves on the complex plane, tells us information about the Taylor series, even if we wish to consider only real functions. Example: Show that \(f(z)=z^2+(3+i)z\) satisfies the Cauchy-Riemann equations. Solution: Let \(z=x+iy\) and \(f(x,y)=u(x,y)+iv(x,y)\) \[\begin{align} f(x,y)&=(x+iy)^2+(3+i)(x+iy)\\ &=x^2-y^2+2ixy+3x+3iy+ix-y \end{align}\] Therefore \[\begin{align} &u(z)=x^2-y^2+3x-y\\ &v(z)=2xy+3y+x \end{align}\] And \[ \frac{\partial u}{\partial x}=2x+3\qquad\frac{\partial u}{\partial y}=-2y-1 \] \[ \frac{\partial v}{\partial x}=-2y+1\qquad\frac{\partial v}{\partial y}=2x+3 \] Both C-R equations are satisfied.