If $\frac{(x+1+i)^n -(x+1-i)^n}{2i}=k$ and $\cot(Q)=x+1$ prove $k=\csc^n(Q)\cdot\sin(nQ)$

Question

If $$\frac{(x+1+i)^n -(x+1-i)^n}{2i}=k$$ and $$\cot(Q)=x+1$$ prove that $$k=\csc^n(Q)\cdot\sin(nQ)$$ replacing value of $\cot{Q}$ $$\frac{(\cot(Q)+i)^n -(\cot(Q)-i)^n}{2i}$$ and taking out $\sqrt{\cot^2(Q)+1}$ as common $$\frac{\csc^n(Q)(\cos(Q)+i/\csc(Q))^n -\csc^n(Q)(\cot(Q)-i/\csc(Q))^n}{2i}$$ finally it becomes $$\frac{\csc^n(Q)(2i/\csc(Q))^n}{2i}$$ applying de-moivre's theorem it becomes $$k=\csc^n(Q).\sin(nQ).(2i)^{n-1}$$ where did I go wrong?

Answer 1

How did you get the line after "finally it becomes"? Looks like you said $$a^n-b^n=(a-b)^n$$ which of course is not true.

Try this: write $$x+1+i=re^{i\theta}\ .$$ Then $$\cot\theta=x+1\ ,\quad r=\sqrt{(x+1)^2+1}=\csc\theta$$ and your expression is $${\rm Im}\bigl((x+1+i)^n\bigr)={\rm Im}\bigl(r^ne^{ni\theta}\bigr)=\csc^n\theta\sin n\theta\ .$$

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Note that since ${\rm Im}(x+1+i)>0$ we have $\csc\theta>0$, otherwise we would have to write $r=|\csc\theta|$.