How do you evaluate the square and products of 5e^(3(pi)i)/4 in polar form?

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Homework Statement


Evaluate the square 0f 5e^(3(pi)i)/4 without using Cartesian form, and also the three different products.


Homework Equations


e^i(theta) = cos(theta) + isin(theta)?


The Attempt at a Solution


I have absolutely no idea here, nothing in my notes even begins to suggest how I can answer this.
 
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All you really need to know is the equation you listed. The square of e^x is (e^x)^2=e^(2x), use this information to solve your problem.
 
yes but what do they mean by the 3 products?
 
With not using the three products I would guess they mean to not use:

<br /> (\cos x+i\sin x)^2=\cos^2x-\sin^2 x+2i\cos x \sin x<br />
 
it says to evaluate the three products of the complex number
 
Then just remove the 'nots' in my previous post.
 
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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