Complex q (2 Viewers)

Luukas.2 · Feb 13, 2024

Here is a more exam-like structuring of question 1, modified so that the result to prove is actually correct:

$\text{Let}\ \ z = 1 + \cos\theta + i\sin\theta$

$\text{(a)} \qquad \text{Show that } |z| = 2\left|\cos{\frac{\theta}{2}}\right|.$

$\text{(b)} \qquad \text{Express $z$ in modulus-argument form.}$

$\text{(c)} \qquad \text{Hence, show that } \left(1 + \cos\theta + i\sin\theta\right)^k + \left(1 + \cos\theta - i\sin\theta\right)^k = 2^{k+1}\left|\cos^k{\frac{\theta}{2}}\right|\cos{\frac{k\theta}{2}}.$

Brief Solution (a)

$\begin{align*} |z| &= \left|(1 + \cos\theta) + i\sin\theta\right| \\ |z|^2 &= (1 + \cos\theta)^2 + \sin^2{\theta} \\ &= 2 + 2\cos\theta = 2 + 2\left(2\cos^2{\frac{\theta}{2}} - 1\right) = 4\cos^2{\frac{\theta}{2}} \\ |z| &= 2\left|\cos{\frac{\theta}{2}}\right| \end{align*}$

Brief Solution (b)

$\text{Since $\cos\theta + i\sin\theta$ lies on the unit circle, $\Re{(z)} \ge 0$, from which it follows that $\alpha = \arg{z} \in \left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$:$

$\begin{align*} \arg{z} &= \tan^{-1}{\frac{\Im{(z)}}{\Re{(z)}}} \\ \tan\alpha &= \frac{\sin\theta}{1 + \cos\theta} = \frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{1 + \left(2\cos^2{\frac{\theta}{2}} - 1\right)} = \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}} \\ \tan\alpha &= \tan{\frac{\theta}{2}} \\ \alpha = \arg{z} &= \frac{\theta}{2} \\ \\ \text{Thus,} \qquad z &= 2\left|\cos{\frac{\theta}{2}}\right|\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right) \end{align*}$

Brief Solution (c)

$\begin{align*} z^k &= \left[2\left|\cos{\frac{\theta}{2}}\right|\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right)\right]^k = 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right)^k \\ &= 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{k\theta}{2}} + i\sin{\frac{k\theta}{2}}\right) \qquad \text{on applying De Moivre's Theorem} \\ \\ \left(\bar{z}\right)^k &= (1 + \cos\theta - i\sin\theta)^k \\ &= \left[2\left|\cos{\frac{\theta}{2}}\right|\left(\cos{\frac{\theta}{2}} - i\sin{\frac{\theta}{2}}\right)\right]^k = 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{\theta}{2}} - i\sin{\frac{\theta}{2}}\right)^k \\ &= 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{k\theta}{2}} - i\sin{\frac{k\theta}{2}}\right) \qquad \text{on applying De Moivre's Theorem} \end{align*}$

$\begin{align*} \text{Hence,} \qquad (1 + \cos\theta + i\sin\theta)^k + (1 + \cos\theta - i\sin\theta)^k &= z^k + \left(\bar{z}\right)^k \\ &= 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{k\theta}{2}} + i\sin{\frac{k\theta}{2}}\right) + 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{k\theta}{2}} - i\sin{\frac{k\theta}{2}}\right) \\ &= 2^k\left|\cos{\frac{\theta}{2}}\right|^k\left(\cos{\frac{k\theta}{2}} + i\sin{\frac{k\theta}{2}} +\cos{\frac{k\theta}{2}} - i\sin{\frac{k\theta}{2}}\right) \\ &= 2^k\left|\cos{\frac{\theta}{2}}\right|^k \times 2\cos{\frac{k\theta}{2}} \\ &= 2^{k+1}\left|\cos^k{\frac{\theta}{2}}\right|\cos{\frac{k\theta}{2}} \qquad \text{as required} \end{align*}$

Complex q (2 Viewers)

Luukas.2

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