Functions of a complex variable
A complex function is not plotted as a curve: you show what it turns each figure of the plane into. That is how you see where the Smith chart comes from, and why the complex exponential describes everything that oscillates and dies out.
01A function that cannot be graphed
A real function \( y = f(x) \) is drawn as a curve: two axes are enough. A complex function \( w = f(z) \) takes a point of the plane and returns another point of the plane: it would take four dimensions to graph it.
The way out is to draw two planes, the starting one and the arrival one, and show what each figure turns into. That is called a mapping, and it is the natural way to understand what a complex function does.
The transfer function of a circuit is a complex function: to each value of \( s = \sigma + j\omega \) it assigns a complex number \( H(s) \). When you run along the imaginary axis you get the frequency response; when you look for the points where \( H \) becomes infinite, the poles appear. Everything that follows in the course is reading a complex function somewhere in the plane.
02The functions you need to know
| Function | What it does to the plane | Where it appears |
|---|---|---|
| \( w = z + c \) | Translates | DC level shift |
| \( w = az \) | Scales and rotates | Gain and phase shift of a stage |
| \( w = z^2 \) | Doubles the angle, squares the modulus | Mixer, second harmonic |
| \( w = 1/z \) | Inverts with respect to the unit circle | Impedance to admittance |
| \( w = e^z \) | Takes horizontal lines to rays and vertical lines to circles | Complex exponential, phasors |
| \( w = \dfrac{z-1}{z+1} \) | Takes the right half-plane to the unit disk | The Smith chart |
The last row deserves attention: that transformation, called the bilinear or Möbius transformation, is the one that turns the half-plane of impedances —infinite and impossible to draw— into a circle of radius one that fits on a sheet of paper. That, and nothing else, is the Smith chart.
On the left, a grid in the z-plane; on the right, what it turns into when the function is applied. Click or tap the left plane to follow a point.
03The complex exponential
It is the most important function of the whole course. Separating modulus and phase:
- The real part of \( z \) controls how much it grows or dies out.
- The imaginary part controls how much it rotates.
- It is periodic with period \( 2\pi j \): \( e^{z + 2\pi j} = e^{z} \). A complex function that is periodic in the imaginary direction, something with no real analog.
When \( e^{st} \) with \( s = \sigma + j\omega \) shows up in the course, everything has already been said: the signal rotates at \( \omega \) radians per second while its amplitude is multiplied by \( e^{\sigma t} \). With \( \sigma \lt 0 \) it dies out, with \( \sigma \gt 0 \) it grows, and with \( \sigma = 0 \) what remains is a pure sinusoid. That single formula contains the whole natural response of a circuit.
Since the exponential is periodic, its inverse cannot be an ordinary function: \( \ln z = \ln|z| + j(\arg z + 2\pi k) \) for any integer \( k \). A branch is chosen —normally the one with \( -\pi \lt \arg z \le \pi \)— and you have to state it. That same phenomenon is what makes the phase measured by an analyzer appear “wrapped” between −180° and 180°, so that it has to be unwrapped.
04Conformal mapping
An analytic function —differentiable in a neighborhood, which is studied in the next topic— has a remarkable geometric property: wherever its derivative does not vanish, it preserves angles. Two curves that crossed at 90° in the z-plane still cross at 90° after being transformed, even though the curves have been completely deformed.
In the lab you can see it for yourself: the starting grid has all its crossings at right angles, and the transformed one, however twisted it ends up, keeps them. Where that breaks down —at \( z = 0 \) for \( w = z^2 \), for example— is exactly where the derivative vanishes, and those points are called critical.
| Used for | How |
|---|---|
| The Smith chart | Taking the half-plane of impedances to a circle |
| Fields and capacitances | Transforming a difficult geometry into a simple one, solving there and coming back |
| Transmission lines | Moving between reflection coefficient and impedance |
| Stability | Mapping the Nyquist contour and counting encirclements of −1 |
05In the lab
Take five normalized impedances —0, 1, ∞, j and −j— and compute \( \Gamma = (z-1)/(z+1) \) for each one. Locate the results on the chart and check that they fall where they should: center, right edge, left edge and the two extremes of the circle.
Draw on paper a 1×1 grid in the first quadrant and transform it with \( w = z^2 \) point by point. Verify that the crossings are still at right angles except at the origin.
On an RC circuit with a step input, capture the response and fit \( e^{\sigma t} \). Check that \( \sigma = -1/RC \) matches the pole of the circuit, which is the point of the plane where the transfer function becomes infinite.
06Common mistakes
- Trying to graph \( w = f(z) \) with a curve. Two planes are needed.
- Forgetting that the logarithm and the root are multivalued and keeping one branch without stating it.
- Believing that \( e^z \) never takes the same value twice. It is periodic in the imaginary direction.
- Assuming that every complex function preserves angles. Only analytic ones do, and only where the derivative does not vanish.
- Confusing \( |e^z| \) with \( e^{|z|} \). The modulus of \( e^z \) is \( e^{x} \), and it depends only on the real part.
07Self-assessment
What is \( |e^{2+j5}| \)?
\( e^2 \approx 7.39 \): the modulus depends only on the real part; the imaginary part only rotates.
What does the imaginary axis turn into under \( w = (z-1)/(z+1) \)?
The unit circle. That is why purely reactive impedances fall on the edge of the Smith chart: they reflect all the power.
What does \( w = z^2 \) do to the first quadrant?
It stretches it until it covers the entire upper half-plane: it doubles all the angles.
Why does \( \ln z \) have infinitely many values?
Because \( e^z \) is periodic with period \( 2\pi j \), so infinitely many values of \( z \) give the same exponential.
If \( s = -100 + j2000 \), what does the signal \( e^{st} \) look like?
An oscillation of 2000 rad/s —about 318 Hz— whose amplitude decays with a time constant \( 1/100 = 10 \) ms. It is the typical response of an underdamped RLC circuit.
What does it mean for a mapping to be conformal?
That it preserves the angles between curves. This is the case for analytic functions wherever the derivative does not vanish.
08Further reading
- Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. The chapters on elementary functions and transformations, with many worked mappings.
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The classic treatment of conformal mapping and its applications to fields.
- David M. Pozar. Microwave Engineering. 4th ed., Wiley, 2011. The bilinear transformation put to work: the Smith chart, derived step by step.