MHB Is it possible to avoid using the complex logarithm?

[aruwin]

Hello.
Can someone please check if I did this correctly. Question and my attempt are as the attached.

 

[DreamWeaver]

Looks alright to me, but in future - no pressure, mind, just a suggestion - if you were to take the time to learn a little LaTex and write it so, you'd probably get a faster response on here. General rule of thumb: make it easier for others to read, and you'll get more and quicker replies. (Sun)

All the best!

Gethin :D

 

[Prove It]

I try to avoid the complex logarithm where possible. Since you know that $\displaystyle \begin{align*} e^{i \,z} = 2 \pm \sqrt{3} \end{align*}$, a real number, that means z must be purely imaginary, and you can solve as you would real numbers, just remembering that your value of z will be periodic every $\displaystyle \begin{align*} 2\pi \end{align*}$ units. So

$\displaystyle \begin{align*} e^{i \,z} &= 2\pm \sqrt{3} \\ i \,z &= \ln{ \left( 2 \pm \sqrt{3} \right) } \\ -z &= i\ln{ \left( 2 \pm \sqrt{3} \right) } \\ z &= -i\ln{ \left( 2 \pm \sqrt{3} \right) } \end{align*}$

and remembering that it is multivalued, that means the complete solution is

$\displaystyle \begin{align*} z = -i\ln{ \left( 2 \pm \sqrt{3} \right) } + 2\pi n \textrm{ where } n \in \mathbf{Z} \end{align*}$

 

[DreamWeaver]

I try to avoid the complex logarithm where possible.

That's pretty much my mantra too. But if in Rome, at least go for the principal value... (Coffee)