# Complex number trig

1. Nov 29, 2005

### square_imp

I need to work out both cos and sine of (2-i). The answer needs to be in the form x+iy where both x and y are real.

So far I have got:

cos (x) = ( e^ix + e^-ix ) / 2 as a general formula which when I substitute in gives:

0.5e^(2i+1) + 0.5e^(-2i-1)

How do I get this into the correct form? and have I used the correct formula?

Thanks, any help is welcome.

2. Nov 29, 2005

### Galileo

Write down the real and imaginary parts of $\exp(2i+1)$ and $\exp(-2i-1)$ and collect terms.

Yes, the formula for cos(x) is correct.

3. Dec 1, 2005

### square_imp

Thanks for the help, I think my problem is my understanding of complex numbers. How do you split the exp terms into real and imaginary parts? Can you help me with that?

4. Dec 1, 2005

### Galileo

Do you know the definition of the exponent of a complex number? $\exp(z)=\exp(x+iy)$

That would be the most basic thing to start with right? How come you are working with expressions like $\exp(1+2i)$ when you don't even know what it means? (And ofcourse, if you don't know, you should find out).

The complex exponential obeys the familiar rule: $\exp(z_1+z_2)=\exp(z_1)\exp(z_2)$, so $\exp(x+iy)=\exp(x)\exp(iy)$. From Euler's formula: $\exp(iy)=\cos y+i\sin y$ so:

$$e^{x+iy}=e^x(\cos y+i\sin y)$$

Last edited: Dec 1, 2005
5. Dec 2, 2005

### square_imp

Thanks for all the help Galileo, I have had another look over complex numbers and I understand it all better now. My only remaining question is whether Euler's formula for exp(iy)=cosy + isiny is for y in radians or is in degree's? I think it is degrees, but I am not sure.

Thanks again

6. Dec 2, 2005

### NateTG

It's in radians. One way to look at it is:
$$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$\sin(x)=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
$$\cos(x)=\sum_{i=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$$