Derivative using complex exponential

forty
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I'm trying find the 15th derivative of exp[(1 + i(3^.5))theta] with respect to theta

To do this do i need to split it into two exponentials, (e^theta).(e^i(3^.5)theta) ??
 
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What is the derivative of eax with respect to x? What is its second derivative? What is its 15th derivative? Do you see my point?
 
[(1+i(3^.5))^15].e^[(1+i(3^.5))theta]

So i can just apply the normal rules for exponentials ??
 
Yes, 1+i(3^.5) is "just a number". You might want to use De Moivre's formula to caculate (1+ i\sqrt{3})^{15} itself.
 
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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