# cosh (z) = i

• Dec 20th 2010, 04:39 PM
dwsmith
cosh (z) = i
$\displaystyle cosh(z)=\mathbf{i} \ \ z\in\mathbb{C}$

$\displaystyle e^{2z}-2\mathbf{i}e^z+1=0$

$\displaystyle \displaystyle e^z=\frac{2\mathbf{i}\pm\sqrt{-4-4}}{2}=\mathbf{i}(1\pm\sqrt{2})$

$\displaystyle \displaystyle z=ln(\mathbf{i}(1+\sqrt{2}))=ln(\mathbf{i}(1+\sqrt {2}))+\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right) \ \ k\in\mathbb{Z}$

$\displaystyle \displaystyle z=ln(\mathbf{i})+ln(1+\sqrt{2})+\mathbf{i}\left(\f rac{\pi}{2}+2\pi k}\right)=\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)+ln(1+\sqrt{2})+\mathbf{i}\left(\frac{\pi }{2}+2\pi k}\right)$

$\displaystyle \displaystyle =ln(1+\sqrt{2})+2\mathbf{i}\left(\frac{\pi}{2}+2\p i k}\right)$

$\displaystyle \displaystyle z=ln(1+\sqrt{2})+\mathbf{i}({\pi}+4\pi k}) \ \ k\in\mathbb{Z}$

I know there is another solution but my question is about this solution. The book doesn't $\displaystyle \mathbf{i}({\pi}+4\pi k})$ but instead has $\displaystyle \displaystyle \mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)$.

The book, when taking the modulus of z, didn't have the $\displaystyle \mathbf{i}$ which would cause the additionally argument. Whenever I take the modulus, I have the $\displaystyle \mathbf{i}$. How can I get rid of it?
• Dec 20th 2010, 06:10 PM
mr fantastic
Quote:

Originally Posted by dwsmith
$\displaystyle cosh(z)=\mathbf{i} \ \ z\in\mathbb{C}$

$\displaystyle e^{2z}-2\mathbf{i}e^z+1=0$

$\displaystyle \displaystyle e^z=\frac{2\mathbf{i}\pm\sqrt{-4-4}}{2}=\mathbf{i}(1\pm\sqrt{2})$

$\displaystyle \displaystyle z=ln(\mathbf{i}(1+\sqrt{2}))$

$\displaystyle =ln(\mathbf{i}(1+\sqrt{2}))+\mathbf{i}\left(\frac{ \pi}{2}+2\pi k}\right) \ \ k\in\mathbb{Z}$ Mr F says: This is wrong.

$\displaystyle \displaystyle z=ln(\mathbf{i})+ln(1+\sqrt{2})+\mathbf{i}\left(\f rac{\pi}{2}+2\pi k}\right)=\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)+ln(1+\sqrt{2})+\mathbf{i}\left(\frac{\pi }{2}+2\pi k}\right)$

$\displaystyle \displaystyle =ln(1+\sqrt{2})+2\mathbf{i}\left(\frac{\pi}{2}+2\p i k}\right)$

$\displaystyle \displaystyle z=ln(1+\sqrt{2})+\mathbf{i}({\pi}+4\pi k}) \ \ k\in\mathbb{Z}$

I know there is another solution but my question is about this solution. The book doesn't $\displaystyle \mathbf{i}({\pi}+4\pi k})$ but instead has $\displaystyle \displaystyle \mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)$.

The book, when taking the modulus of z, didn't have the $\displaystyle \mathbf{i}$ which would cause the additionally argument. Whenever I take the modulus, I have the $\displaystyle \mathbf{i}$. How can I get rid of it?

You need to review the complex logarithm.
• Dec 20th 2010, 06:18 PM
dwsmith
$\displaystyle z=\mathbf{i}(1+\sqrt{2})$

$\displaystyle ln(z)=ln|z|+\mathbf{i}arg(z)$

$\displaystyle \displaystyle |z|=\mathbf{i}(1+\sqrt{2}) \ \ arg(z)=\frac{\pi}{2}+2\pi k$

I don't know what is wrong.

Oh, $\displaystyle |\mathbf{i}|=1$