# Sketching exponential curves with complex numbers

1. Jan 5, 2012

### dan5

How do you go about sketching y as a function of t for t≥0

y= e(0.5t + i(√7/2)t) - e(0.5t-i(√7/2)t)

I know it goes through the origin, and the gradient is positive here. But i'm unsure on how to deal with the imaginary numbers when I have a graph of y vs t.

2. Jan 5, 2012

### micromass

Can you try to simplify it using the definition of complex exponentiation. That is:

$$e^{a+ib}=e^a(\cos(b)+i\sin(b))$$

3. Jan 5, 2012

### dan5

Thanks for that idea, I have done that and am left with

y=e0.5t ( i 2 sin (√7/2)t )

But as I am meant to be plotting y as a function of t, I don't see how the imaginary part will factor in

4. Jan 5, 2012

### micromass

Well, it will be immediately obvious that no point on the curve will have a real part. So the curve will only move on the imaginary axis.

5. Jan 5, 2012

### HallsofIvy

In general, a function from R to C (real numbers to complex numbers) requires two dimensions for the value as well as one dimension for the argument- in other words, a three dimensional graph! However, as micromass points out, $y= [e^{0.5t}sin(\sqrt{7}{2}t)]i$ is always imaginary so you you really only need a single dimension for that.

Graph $y= e^{0.5t}sin((\sqrt{7}/2)t)$, clearly labeling the "y" axis as imaginary numbers.