Calculating cos(2-i): Solving Complex Trig Formulas

[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.

 

[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.

 

[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?

 

[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:


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

 

[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

 

[NateTG]

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.

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