# Why complex numbers?

1. Mar 3, 2012

### likephysics

Yes, they make things simpler. But how?
I've never come across a comparison of life with complex numbers and without?
Can some one point me to an example or give one. An electrical engineering example would be great.

2. Mar 3, 2012

### sophiecentaur

Did you try a textbook?
It's difficult to answer if we don't know your level. A book on 3 Phase Power may be the thing for you.

3. Mar 3, 2012

### skeptic2

If you have a series network of a 160 ohm resistor and a 1 uF capacitor and you connect it across the output of a 1 Vpp, 1 kHz generator, you will measure 0.7 Vpp across the resistor and 0.7 Vpp across the capacitor.

How would you explain that result without using complex numbers?

4. Mar 3, 2012

### TheShrike

It's not an electricity example, but perhaps the simplest way to see that complex numbers make life easier is that differentiating/integrating complex exponentials is easier than the same with sines and cosines.

5. Mar 3, 2012

### sophiecentaur

You can do all electrical calculations using trig functions because any waveform is Real. Complex numbers make it much easier. That's all.

6. Mar 3, 2012

### Gordianus

Try QM without complex numbers.

7. Mar 3, 2012

### skeptic2

Certainly, but how would you explain 0.7 + 0.7 = 1 without using complex numbers? Yes you can explain it using vectors but vectors really amount to complex numbers.

8. Mar 4, 2012

### Jolb

How about trigonometry?! The trig identities you struggle with in 8th grade are all rendered completely trivial using Euler's formula.

I'd like someone to show me a proof of the double angle formula that doesn't use complex numbers. Here's my (almost) one-line proof:

$$cos(2\theta)=Re[e^{i(2\theta)}]=Re[\left (e^{i\theta} \right )^2]=Re[\left ( cos\theta+isin\theta \right )^2]=Re[cos^2\theta-sin^2\theta+2icos\theta sin\theta]=cos^2\theta-sin^2\theta$$

Notice you get sin2θ for free in the same proof, just by taking the imaginary part.

Deriving these with normal trig is a challenge, I dare you to try it.

Another more advanced example is the use of residue calculus on contour integrals in the complex plane to calculate real integrals. This also saves you a ton of work.

9. Mar 4, 2012

### sophiecentaur

I remember doing multiple angle formulae long before we got into complex numbers. It may not have been pretty but I think I remember it involved drawing triangles and very basic trig - followed by a lot of re-arrangement.
Complex numbers are a fantastically useful tool but I think they relate to basic trig in the same way that Warp travel relates to sub-Warp travel. They take you into another dimension (literally) which makes your journey quicker. But the results always need to be taken in the right context after complex jiggery pokery.
I wouldn't be without 'em personally.

10. Mar 4, 2012

### Dickfore

If you consider the Taylor-MacLaurin series expansion of the rational function:
$$f(x) = \frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}{(-1)^{n} \, x^{2 n}}$$
you will find that the radius of convergence is 1. It seems very puzzling why such a function, which is well behaved everywhere on the real axis, would have a Taylor series with a finite radius of convergence.

The reason is, of course, that, being a meromorphic function, it has poles in the complex plane. For this function, the poles are purely imaginary $z_{1/2} = \pm i$, and are "invisible" on the real-axis. Nevertheless, their distance to the origin is 1, and it determines the radius of convergence for the above Taylor series.

In fact, the theory of analytic functions is the primary gain from studying complex analysis. Analytic functions always have a Taylor series expansion.

11. Mar 4, 2012

### sophiecentaur

Here's a derivation for all combinations of sin/cos (A+/-B) which your granny might understand. It works for A or B in one quadrant and is a lot shorter than I remember (with my see-back-to-17 glasses).