How Do You Calculate the Real and Imaginary Parts of \( e^{e^z} \)?

[Deanmark]

Let f(z) = $e^{e^{z}}$ . Find Re(f) and Im(f).

I don't know how to deal with the exponential within an exponential. Does anybody know how to deal with this?

 

[Euge]

Yes. Recall Euler’s formula: $e^{iy} = \cos y + i\sin y$ for all real $y$. For $z = x + yi$, we would then have $e^z = e^{x + yi} = e^x e^{yi} = e^x(\cos y + i\sin y)$. If $w = e^z$, then $$e^w = e^{e^x\cos y + i(e^x\sin y)} = e^{e^x\cos y}\, e^{i(e^x\sin y)} = \cdots$$ Take it from here.

 

[math771]

To find the real part of f(z), we can use the fact that Re(f) = $e^{Re(z)}\cos(Im(z))$. Since z is a complex number, we can write it as z = x + iy, where x and y are real numbers. Then, using this form for z, we can rewrite f(z) as f(z) = $e^{e^{x+iy}}$. Using the property of exponents, we can write this as f(z) = $e^{e^{x}e^{iy}}$. Now, we can use Euler's formula to write $e^{iy}$ as $\cos(y) + i\sin(y)$. Plugging this into our expression for f(z), we get f(z) = $e^{e^{x}(\cos(y)+i\sin(y))}$. Finally, taking the real part of this expression, we get Re(f) = $e^{e^{x}}\cos(e^{x}\cos(y))$.

To find the imaginary part of f(z), we can use the fact that Im(f) = $e^{Re(z)}\sin(Im(z))$. Using the same steps as above, we can rewrite f(z) as f(z) = $e^{e^{x}(\cos(y)+i\sin(y))}$. Taking the imaginary part of this expression, we get Im(f) = $e^{e^{x}}\sin(e^{x}\cos(y))$.

So, in summary, Re(f) = $e^{e^{x}}\cos(e^{x}\cos(y))$ and Im(f) = $e^{e^{x}}\sin(e^{x}\cos(y))$.