complex number z satisfies exp(z/i)= 2*sin(z). Then the imaginary part of z is...?

[aledg97]

The complex number z satisfies exp(z/i)= 2*sin(z). Then the imaginary part of z is...?

The answer provided is: -1/4 * log2

My attempt was to write sin(z) as (exp(iz)-exp(-iz))/2i
but then I couldn't find the imaginary part, any help please?

 

[Deleted member 4993]

exp(z/i)=sin(z)

Find the imaginary part of z

My attempt was to write sin(z) as (exp(iz)-exp(-iz))/2i
but then I couldn't find the imaginary part, I know that I should get some expression with the logarithm inside, but any help please?

Is that the EXACT wording of the problem?

 

[aledg97]

Is that the EXACT wording of the problem?

No, sorry, I made a mistake. I did the correction.

 

[HallsofIvy]

You have \(\displaystyle e^{z/i}= 2 sin(z)\). Yes, \(\displaystyle sin(z)= \frac{e^{iz}- e^{-iz}}{2i}\) so we can write the equation as \(\displaystyle e^{z/i}= \frac{e^{iz}- e^{-iz}}{i}\).

I would now use the fact that \(\displaystyle \frac{1}{i}= -i\) to write that as \(\displaystyle e^{-iz}= \frac{e^{iz}- e^{-iz}}{i}\). Multiply both sides by i: \(\displaystyle ie^{-iz}= e^{iz}- e^{-iz}\). Add \(\displaystyle e^{-iz}\) to both sides: \(\displaystyle e^{iz}= (1+ i)e^{-iz}\).


Multiply both sides by \(\displaystyle e^{iz}\): \(\displaystyle e^{2iz}= (1+ i)\). Let \(\displaystyle y= e^{iz}\) and the equation becomes \(\displaystyle y^2= 1+ i\). \(\displaystyle y= e^{iz}= \pm\sqrt{1+ i}\).

Can you finish that?

 

[aledg97]

You have \(\displaystyle e^{z/i}= 2 sin(z)\). Yes, \(\displaystyle sin(z)= \frac{e^{iz}- e^{-iz}}{2i}\) so we can write the equation as \(\displaystyle e^{z/i}= \frac{e^{iz}- e^{-iz}}{i}\).

I would now use the fact that \(\displaystyle \frac{1}{i}= -i\) to write that as \(\displaystyle e^{-iz}= \frac{e^{iz}- e^{-iz}}{i}\). Multiply both sides by i: \(\displaystyle ie^{-iz}= e^{iz}- e^{-iz}\). Add \(\displaystyle e^{-iz}\) to both sides: \(\displaystyle e^{iz}= (1+ i)e^{-iz}\).


Multiply both sides by \(\displaystyle e^{iz}\): \(\displaystyle e^{2iz}= (1+ i)\). Let \(\displaystyle y= e^{iz}\) and the equation becomes \(\displaystyle y^2= 1+ i\). \(\displaystyle y= e^{iz}= \pm\sqrt{1+ i}\).

Can you finish that?

I am not sure I can. What I would do is:

exp(2iz)=1+i ; exp(-2y+2ix)=1+i; -2y + 2ix= log(1+i) doesn't allow me to take the imaginary part. So maybe taking the modulus? |exp(-2y+2ix)| = sqrt(2); |exp(-2y)|=sqrt(2); and continue, but I think there must be a more logical and safe approach

 

[HallsofIvy]

The equation has been reduced to \(\displaystyle e^{iz}= \pm\sqrt{1+ i}\). First, look at \(\displaystyle e^{iz}= \sqrt{1+ i}\). Since the left had side is written as an exponential, write the right hand side as an exponential: 1+ i has absolute value \(\displaystyle \sqrt{1^2+ 1^2}= \sqrt{2}\) and argument \(\displaystyle arctan(1/1)= arctan(1)= \pi/4\). So \(\displaystyle 1+ i= \sqrt{2}e^{i\pi/4}\). The equation is \(\displaystyle e^{iz}= \sqrt{2}e^{i\pi/4}\). Taking the logarithm of both sides, \(\displaystyle iz= log(\sqrt{2})+ i\pi/4\)\(\displaystyle = \frac{1}{2} log(2)+ \left(\frac{\pi}{4}\right)i\). Finally, divide both sides by i: \(\displaystyle z= \frac{1}{2i}log(2)+ \frac{\pi}{4}= \frac{\pi}{4}- \frac{1}{2}log(2)i\).