# Euler's Formula - Need help!

1. Dec 24, 2003

### AndersHermansson

I just started studying complex numbers. It says complex numbers in polar form can be expressed as a power of e:

e^ix = cosx + isinx

I don't quite understand how this equation works.
How do i evaluate e^ix? And how does taking e to the power of ix get me a complex number a + bi or even in polar form?

2. Dec 24, 2003

### NateTG

To get all complex numbers, you need the somewhat more general form:
$$re^{i\theta}=r\sin{\theta}+ir\cos{\theta}$$

Now, the other direction is also not bad
$$z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}$$

If you think of $$r$$ as the radius, and $$\theta$$ as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.

3. Dec 24, 2003

### AndersHermansson

So $$re^{i\theta}$$ is merely a convenience in writing?

4. Dec 24, 2003

### mathman

It is more than a convenience. One way to look at it is by power series of eix and compare to power series for sinx and cos x. You will see that eix=cosx+isinx.

5. Dec 24, 2003

### AndersHermansson

Ok, yes i think i'm beginning to understand. How do one evaluate xi? For example, what's 32i?

6. Dec 24, 2003

### krab

If you are thinking of x^n being the product of n x's, and want to apply that here, you are out of luck. When the exponent is imaginary, it doesn't really carry over. Think of it as a generalization.

Are you familiar with Taylor series?

$$e^x=1+x+{x^2\over 2}+{x^3\over 3!}+{x^4\over 4!}+...$$

$$\sin x=x-{x^3\over 3!}+{x^5\over 5!}-...$$

$$\cos x=1-{x^2\over 2}+{x^4\over 4!}-...$$

The result follows from these.

7. Dec 24, 2003

### NateTG

Well, you can do the following:

$$3^{2i}=e^{\ln{3} *2i}=\cos{2\ln{3}}+i\sin{2\ln{3}}$$

If you're feeling a bit more adventerous, there is a power series for $$3^x$$ which is going to give you the same result.

8. Dec 24, 2003

### AndersHermansson

Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.

9. Dec 24, 2003

### Hurkyl

Staff Emeritus
For the record, x^y is usually a multivalued function.

10. Dec 24, 2003

### NateTG

Think of it as a way to calculate $$x^{iy}$$.

11. Dec 25, 2003

### phoenixthoth

should that be arctan(y/x)?