Find the real and imaginary part of z

Question

let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$

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Rationalizing the denominator: $$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$

$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$

thus

$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$

$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$

According to the binomial theorem,

$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$

we get

$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$

...and that is where I'm stuck. What do you think? Thanks for the attention.

Answer 1

The denominator is $2+2\sin\theta$, the numerator is, setting $2\alpha=\theta$, $$ 1+e^{2i\alpha}=2e^{i\alpha}\cos\alpha $$ Then your number is $$ \left(\frac{\cos\alpha}{1+\sin2\alpha}\right)^{n}e^{ni\alpha} $$

Answer 2

HINT: Express the fraction as $r e^{i\theta}$ and compute $r^n e^{i n\theta}$.

Answer 3

we know that $a+ib=\sqrt{a^{2}+b^{2}}\cdot e^{i\arctan \frac{b}{a}}\Rightarrow \frac{a+ib}{a-ib}=e^{2i\arctan \frac{b}{a}}$ and $\cos \theta +i\sin \theta =e^{i\theta }$ then:

$\left( \frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } \right)^{n}=e^{i\cdot 2n\arctan \frac{\cos \theta }{1+\sin \theta }}=\cos \left( 2n\arctan \frac{\cos \theta }{1+\sin \theta } \right)+i\sin \left( 2n\arctan \frac{\cos \theta }{1+\sin \theta } \right)$

Answer 4

Noting $$ 1+\sin\theta=1+\cos(\frac{\pi}{2}-\theta)=2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}), \cos\theta=\sin (\frac{\pi}{2}-\theta)=2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})$$ one has \begin{eqnarray} z&=&\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n\\ &=&\left( \frac{1+\cos(\frac{\pi}{2}-\theta) + i\sin (\frac{\pi}{2}-\theta)}{1+\cos(\frac{\pi}{2}-\theta) - i\sin (\frac{\pi}{2}-\theta)} \right)^n\\ &=&\left( \frac{2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}) + i2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})}{2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}) - i2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})} \right)^n\\ &=&\left( \frac{\cos(\frac{\pi}{4}-\frac{\theta}{2}) + i\sin(\frac{\pi}{4}-\frac{\theta}{2})}{\cos(\frac{\pi}{4}-\frac{\theta}{2}) - i\sin(\frac{\pi}{4}-\frac{\theta}{2})} \right)^n\\ &=&(\cos(\frac{\pi}{2}-\theta) + i\sin(\frac{\pi}{2}-\theta))^n\\ &=&\cos(\frac{n\pi}{2}-n\theta) + i\sin(\frac{n\pi}{2}-n\theta) \end{eqnarray} and hence the real and imaginary parts are easy to get.

Answer 5

$$\left( \frac { 1+\sin \theta +i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) ^{ n }={ \left( 1+\frac { 2i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) }^{ n }=\\ ={ \left( 1+\frac { 2i\cos \theta \left( 1+\sin \theta +i\cos \theta \right) }{ \left( 1+\sin \theta -i\cos \theta \right) \left( 1+\sin \theta +i\cos \theta \right) } \right) }^{ n }=\\ ={ \left( 1+\frac { 2i\cos \theta +2isin{ \theta \cos { \theta } }{ -2\cos ^{ 2 }{ \theta } } }{ 2+2\sin \theta } \right) }^{ n }=\\ ={ \left( 1+\frac { i\cos \theta +isin{ \theta \cos { \theta } }{ -\cos ^{ 2 }{ \theta } } }{ 1+\sin \theta } \right) }^{ n }=\\ ={ \left( 1-\frac { \cos ^{ 2 }{ \theta } }{ 1+sin{ \theta } } +i\frac { \cos \theta +sin{ \theta \cos { \theta } } }{ 1+\sin \theta } \right) }^{ n }={ \left( \sin { \theta } +i\cos { \theta } \right) }^{ n }=\\ ={ \left( \cos { \left( \frac { \pi }{ 2 } -\theta \right) +i\sin { \left( \frac { \pi }{ 2 } -\theta \right) } } \right) }^{ n }=\cos { n\left( \frac { \pi }{ 2 } -\theta \right) +i\sin { \left( n\left( \frac { \pi }{ 2 } -\theta \right) \right) } } \\ \\ \\ \\ $$

Answer 6

Let's get a picture:

$|1 + \sin\theta + i\cos\theta| = |1 + \sin\theta - i\cos\theta|$

$\left|\frac {1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\right| = 1$

When we divide complex numbers the argument of the ratio equals the difference between the arguments.

let $\phi = \frac {\pi}{2} - \theta$

$1 + \sin\theta + i\cos\theta = 1 + \cos\phi + i\sin\phi$

$\frac {1 + \cos \phi + i\sin\theta}{1 + \cos\phi - i\sin\phi} = e^{\phi i}$

$\left(\frac {1 + \cos \phi + i\sin\phi}{1 + \cos\phi - i\sin\phi}\right)^n = e^{n\phi i}$

or

$e^{n(\frac \pi 2 - \theta) i}$

Answer 7

Following your rationalizing the denominator method, you err when expanding the square. You should get $$ \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} = \frac{(1+\sin\theta)^2 - (\cos\theta)^2 + 2i\cos\theta(1+\sin\theta)}{(1+\sin\theta)^2 + (\cos\theta)^2} $$ This can then be simplified by expanding the $(1+\sin\theta)^2$ term and using $(\sin\theta)^2 + (\cos\theta)^2 = 1$. $$ \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} = \frac{1 - (\cos\theta)^2 + (\sin\theta)^2 + 2\sin\theta + 2i\cos\theta(1+\sin\theta)}{1+2\sin\theta + (\sin\theta)^2 + (\cos\theta)^2} = \frac{2\sin\theta(1+\sin\theta) + 2i\cos\theta(1+\sin\theta)}{2 + 2\sin\theta} = \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta) $$ It should be clear then that $$ \left(\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\right)^n = i^n\left[\cos(n\theta)-i\sin(n\theta) \right] $$ and you should be able to take real and imaginary parts from there.

Answer 8

It is convenient to use $e^{i\theta}=\cos\theta + i\sin\theta$.

We obtain \begin{align*} \color{blue}{z}&\color{blue}{=\left(\frac{1+\sin \theta +i\cos \theta}{1+\sin\theta-i\cos \theta}\right)^n}\\ &=\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n\\ &=\left(\frac{1+ie^{-i\theta}}{-ie^{i\theta}(1+e^{-i\theta})}\right)^n\\ &=\left(ie^{-i\theta}\right)^n\\ &=i^ne^{-in\theta}\\ &=\color{blue}{i^n\left(\cos (n\theta) - i\sin(n\theta)\right)} \end{align*}