# Help with Complex Number Question (de Moivre's Theorem) (1 Viewer)

#### policy

##### New Member
The roots of the equation t²-2t+2=0 are a and b. Prove that [(x+a)ⁿ-(x+b)ⁿ]/(a-b) = (sin nθ)/(sinⁿ θ), where cot θ = x +1

Any help would be appreciated!

#### fan96

##### 617 pages
$\cot \theta = x + 1 \implies \cos \theta = (x+1)\sin\theta$

Note that $\cot \theta \in \mathbb{R} \implies x \in \mathbb{R}$

So

${\rm cis\,} \theta = \sin \theta(x + 1) + i \sin \theta = (x+a) \sin \theta$

And

$\sin n \theta = \mathrm{Im} ({\rm cis \,}^n \theta)$

So we have

$\frac{\sin n\theta}{\sin^n\theta} = \frac{{\rm Im\,}[(x+a)^n \sin^n \theta]}{\sin^n \theta} = {\rm Im \,}[(x+a)^n]$

So now all you need to prove is

$\frac{ {\rm Re \,}[(x+a)^n] + i \cdot {\rm Im \,}[(x+a)^n] - (x+b)^n}{a-b} = {\rm Im \,}[(x+a)^n]$

(Start by finding $a-b$)

EDIT: this question was also discussed here: http://community.boredofstudies.org/14/mathematics-extension-2/368909/hsc-2018-mx2-marathon-4.html

EDIT #2: Here's the rest:

$a = 1+i$

$b = 1-i$

$a - b = 2i$

We need to prove

\begin{aligned} &\iff {\rm Re \,}[(x+a)^n] + i \cdot {\rm Im \,}[(x+a)^n] - (x+b)^n} = 2i \cdot {\rm Im \,}[(x+a)^n] \\ &\iff {\rm Re \,}[(x+a)^n] - i \cdot {\rm Im \,}[(x+a)^n] = (x+b)^n \\ &\iff \overline{(x+a)^n} = (x+b)^n \\ &\iff (x + \bar{a})^n = (x+b)^n \quad (\because x \in \mathbb{R})\end{aligned}

And since $a = \overline b$ the proof is complete.

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#### Q16slayer

##### New Member
Assume, without loss of generality (because you can switch a and b and the expression would still be the same)
a = 1 + i
b = 1 - i
(worked out by quadratic formula)
x + 1 = cot theta
sub a, b, x+1 basically everything into LHS expression and simplify

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