Question

$\left( \frac{cos\theta + isin\theta}{sin\theta + icos\theta} \right)^4$

Answer 1

$cos(8\theta) + isin(8\theta)$

Answer 2

The problem asks us to simplify the given complex number expression: $\left( \frac{cos\theta + isin\theta}{sin\theta + icos\theta} \right)^4$

We will simplify the expression inside the parenthesis first, then apply the power of 4.

Step 1: Express the numerator in polar form. The numerator is $cos\theta + isin\theta$. This is already in Euler's form: $\text{Numerator} = e^{i\theta}$

Step 2: Express the denominator in polar form. The denominator is $sin\theta + icos\theta$. We can use the trigonometric identities: $sin\theta = cos\left(\frac{\pi}{2} - \theta\right)$ $cos\theta = sin\left(\frac{\pi}{2} - \theta\right)$ Substituting these into the denominator: $\text{Denominator} = cos\left(\frac{\pi}{2} - \theta\right) + isin\left(\frac{\pi}{2} - \theta\right)$ This is in Euler's form: $\text{Denominator} = e^{i\left(\frac{\pi}{2} - \theta\right)}$

Step 3: Simplify the fraction inside the parenthesis. Let $Z$ be the expression inside the parenthesis: $Z = \frac{cos\theta + isin\theta}{sin\theta + icos\theta} = \frac{e^{i\theta}}{e^{i\left(\frac{\pi}{2} - \theta\right)}}$ Using the property $\frac{e^{ia}}{e^{ib}} = e^{i(a-b)}$: $Z = e^{i\left(\theta - \left(\frac{\pi}{2} - \theta\right)\right)} = e^{i\left(\theta - \frac{\pi}{2} + \theta\right)} = e^{i\left(2\theta - \frac{\pi}{2}\right)}$

Step 4: Apply the power of 4 to the simplified fraction. The given expression is $Z^4$: $Z^4 = \left( e^{i\left(2\theta - \frac{\pi}{2}\right)} \right)^4$ Using De Moivre's Theorem, $(e^{ix})^n = e^{inx}$: $Z^4 = e^{i \cdot 4\left(2\theta - \frac{\pi}{2}\right)} = e^{i(8\theta - 2\pi)}$

Step 5: Convert the result back to trigonometric form and simplify. Using Euler's formula, $e^{ix} = cos x + isin x$: $Z^4 = cos(8\theta - 2\pi) + isin(8\theta - 2\pi)$ Since $cos(x - 2\pi) = cos x$ and $sin(x - 2\pi) = sin x$ (due to the periodicity of trigonometric functions): $Z^4 = cos(8\theta) + isin(8\theta)$

The simplified expression is $cos(8\theta) + isin(8\theta)$.

Explanation of the solution:

1. Convert the numerator $cos\theta + isin\theta$ to its Euler form $e^{i\theta}$.
2. Convert the denominator $sin\theta + icos\theta$ to its Euler form $e^{i(\frac{\pi}{2} - \theta)}$ by using $sin\theta = cos(\frac{\pi}{2} - \theta)$ and $cos\theta = sin(\frac{\pi}{2} - \theta)$.
3. Divide the Euler forms: $\frac{e^{i\theta}}{e^{i(\frac{\pi}{2} - \theta)}} = e^{i(\theta - (\frac{\pi}{2} - \theta))} = e^{i(2\theta - \frac{\pi}{2})}$.
4. Raise the result to the power of 4 using De Moivre's theorem: $(e^{i(2\theta - \frac{\pi}{2})})^4 = e^{i4(2\theta - \frac{\pi}{2})} = e^{i(8\theta - 2\pi)}$.
5. Convert back to trigonometric form: $cos(8\theta - 2\pi) + isin(8\theta - 2\pi)$.
6. Use the periodicity of cosine and sine functions ($cos(x-2\pi) = cos x$, $sin(x-2\pi) = sin x$) to get $cos(8\theta) + isin(8\theta)$.