Simplifying Complex Arithmetic: Converting cos(1+i) to Cartesian and Euler Forms

[PhysicsMark]

Homework Statement

I have a very simple complex arithmetic question.

How do I express the quantity cos(1+i) in Cartesian (a+ib) and Euler(re^i*theta)

Is this the right track?:

cos(1+i)={e^{i(1+i)}\over2}+{e^{-i(1+i)}\over2}

I know that:

cos(1+i)=cos(1)cos(i)-sin(1)sin(i)

but I am not sure how to get the "i's" out of the cosine and sine.

 

[Dick]

Well, sure you are on the right track. The same formula you used before. cos(i)=(e^(i*i)+e^(i*(-i))/2=(e^(-1)+e^1)/2. That's cosh(1). Use similar expression for sin(i).

 

[PhysicsMark]

Well, sure you are on the right track. The same formula you used before. cos(i)=(e^(i*i)+e^(i*(-i))/2=(e^(-1)+e^1)/2. That's cosh(1). Use similar expression for sin(i).

So then as far as cartesian form goes:


cos(1+i)=cos(1)cosh(1)-i(sin(1)sinh(1))

If that is correct, what is the relationship between the hyperbolic functions and the notation:

re^{i\theta}

 

[Dick]

Now you've got cos(1+i) written in the form a+ib where a and b are real. Now it's the usual relation with the polar form. r=sqrt(a^2+b^2), theta=arctan(b/a), right? There's nothing terribly spooky or mysterious going on here.

 

[PhysicsMark]

Thanks Dick. I don't know why this stuff cooks my noodle so much. I feel as though I cannot wrap my head around it. I always think there is some Identity that I am not taking into account or that my answers are not nearly simplified enough.

One last question. Would you consider it correct to write the equation in Euler form this way:


\sqrt{(cos(1)cosh(1))^2+(sin(1)sinh(1))^2}e^{i(arctan{(sin(1)sinh(1)\over(cos(1)cosh(1)})}

 

[Dick]

That's about as good as I could come up with. But you are missing a sign in the argument. You could also write tan(1)*tanh(1) in the argument. But that's not all that much of a real simplification, is it?