How Do You Convert sin^2(z) into x+iy Form?

jjangub
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Homework Statement


How do I change sin^2(z) to x+iy form? (z=x+iy)
I have to put this x and y to arctan(y/x)

Homework Equations


The Attempt at a Solution


I tried to use sin^2(z) = 1/2 -1/2(cos(2z)) or sin(z) = ((e^(iz) - e^(-iz))/2i)^2
but both ways I cannot take out i.
Or isn't the sin only takes the imagenary part? so there is no x and only y exists.
It brings us that artan(0) = 0?
I am confused about these cases like when there is sin(z) and cos(z)...
Thank you.
 
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z is real?
 
How would you expand Sin(x + i y)?
Using a trig identity.
 
jjangub said:

Homework Statement


How do I change sin^2(z) to x+iy form? (z=x+iy)
I have to put this x and y to arctan(y/x)

Homework Equations


The Attempt at a Solution


I tried to use sin^2(z) = 1/2 -1/2(cos(2z)) or sin(z) = ((e^(iz) - e^(-iz))/2i)^2
but both ways I cannot take out i.
Or isn't the sin only takes the imagenary part? so there is no x and only y exists.
It brings us that artan(0) = 0?
I am confused about these cases like when there is sin(z) and cos(z)...
Thank you.

that's not how it works
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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