Problems on Modulus-Amplitude form

\(\bf z=1+i\cot\theta\) where \(\bf \pi~ \text{\textless}~\theta~\text{\textless} ~\frac{3\pi}{2} \) #

  Problem-1: Find the modulus amplitude form of \(\bf z=1+i\cot\theta\) where \(\bf \pi~ \text{\textless}~\theta~\text{\textless} ~\frac{3\pi}{2} \) .\( \\ \) Solution:

  Given that \(\bf z=1+i\cot\theta\) where \(\bf \pi~ \text{\textless}~\theta~\text{\textless} ~\frac{3\pi}{2} \) .\( \\ \)   Let \(\bf z=r(\cos\alpha +i \sin\alpha)\) where \(\alpha\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1+i\cot\theta =r(\cos\alpha +i \sin\alpha) \nonumber\\ \implies &\bf 1=r\cos\alpha \nonumber\\ \bf\bf~and~&\bf \cot\theta=r\sin\alpha\ \tag{1} \end{align}  squaring and adding \begin{align} &\bf r^{2}=\cosec^{2}\theta\nonumber\\ \implies &\bf r=- \cosec\theta\tag{2}\\ \end{align}  Since \(\bf \cosec\theta~\text{\textless}~0\) when \(\bf \pi~ \text{\textless}~\theta~\text{\textless} ~\frac{3\pi}{2} \). Now putting (2) in (1), we get \( \\ \) \begin{align} &\bf \cos\alpha=-\sin\theta\nonumber\\ \bf~and~&\bf \sin\alpha=-\cos\theta\nonumber\\ \end{align}  then \begin{align} &\bf \alpha=\tan^{-1}{\left|\frac{-\cos\theta}{-\sin\theta}\right|} – \pi\nonumber \end{align}

Since if \(\bf \sin\alpha=a~and~\cos\alpha=b \) and
1. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right| \)
2. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\pi-\tan^{-1}\left|\frac{b}{a}\right|\)
3. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right|-\pi\)
4. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=-\tan^{-1}\left|\frac{b}{a}\right|\)

\begin{align} &\bf \alpha= \tan^{-1}{\left[\cot\theta\right]}- \pi \nonumber\\ \implies &\bf \alpha= \tan^{-1}{\left[\tan\left(\frac{\pi}{2}-\theta\right)\right]}- \pi \nonumber\\ \implies &\bf \alpha= \frac{\pi}{2}-\theta- \pi \nonumber\\ \implies &\bf \alpha= -\left(\frac{\pi}{2}+\theta \right)\nonumber \end{align}   Therefore the modulus amplitude form of \(\bf z=1+i\cot\theta\) is \( \bf 1+i\cot\theta=\cos\left(\frac{\pi}{2}+\theta \right) -i\sin\left(\frac{\pi}{2}+\theta \right) \) where \(\bf \pi~ \text{\textless}~\theta~\text{\textless} ~\frac{3\pi}{2} \) .

\(\bf z=1+i\cot\theta\) where \(\bf 0~ \text{\textless}~\theta~\text{\textless} ~\pi \) #

  Problem-2: Find the modulus amplitude form of \(\bf z=1+i\cot\theta\) where \(\bf 0~ \text{\textless}~\theta~\text{\textless} ~\pi \) .\( \\ \) Solution:

Given that \(\bf z=1+i\cot\theta\) where \(\bf 0~ \text{\textless}~\theta~\text{\textless} ~\pi \) .\( \\ \)   Let \(\bf z=r(\cos\alpha +i \sin\alpha)\) where \(\alpha\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1+i\cot\theta =r(\cos\alpha +i \sin\alpha) \nonumber\\ \implies &\bf 1=r\cos\alpha \nonumber\\ \bf\bf~and~&\bf \cot\theta=r\sin\alpha\ \tag{1} \end{align}  squaring and adding \begin{align} &\bf r^{2}=\cosec^{2}\theta\nonumber\\ \implies &\bf r= \cosec\theta\tag{2}\\ \end{align}  Since \(\bf \cosec\theta~>~0\) when \(\bf 0~ \text{\textless}~\theta~\text{\textless} ~\pi \). Now putting (2) in (1), we get \( \\ \) \begin{align} &\bf \cos\alpha=\sin\theta\nonumber\\ \bf~and~&\bf \sin\alpha=\cos\theta\nonumber\\ \end{align}  then \begin{align} &\bf \alpha=\tan^{-1}{\left|\frac{-\cos\theta}{-\sin\theta}\right|} \nonumber \end{align}

Since if \(\bf \sin\alpha=a~and~\cos\alpha=b \) and
1. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right| \)
2. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\pi-\tan^{-1}\left|\frac{b}{a}\right|\)
3. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right|-\pi\)
4. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=-\tan^{-1}\left|\frac{b}{a}\right|\)

\begin{align} &\bf \alpha=\tan^{-1}{\left[\cot\theta\right]} \nonumber\\ \implies &\bf \alpha=\tan^{-1}{\left[\tan\left(\frac{\pi}{2}-\theta\right)\right]} \nonumber\\ \implies &\bf \alpha= \frac{\pi}{2}-\theta \nonumber \end{align}   Therefore the modulus amplitude form of \(\bf z=1+i\cot\theta\) is \( \bf 1+i\cot\theta=\cos\left(\frac{\pi}{2}-\theta \right) +i\sin\left(\frac{\pi}{2}-\theta \right) \) where \(\bf 0~ \text{\textless}~\theta~\text{\textless} ~\pi \) .

\(\bf z=1+i\cot\theta\) where \(\bf \frac{3\pi}{2}~ \text{\textless}~\theta~\text{\textless} ~2\pi \) #

  Problem-3: Find the modulus amplitude form of \(\bf z=1+i\cot\theta\) where \(\bf \frac{3\pi}{2}~ \text{\textless}~\theta~\text{\textless} ~2\pi \) .\( \\ \) Solution:&

Given that \(\bf z=1+i\cot\theta\) where \(\bf \frac{3\pi}{2}~ \text{\textless}~\theta~\text{\textless} ~2\pi \) .\( \\ \)   Let \(\bf z=r(\cos\alpha +i \sin\alpha)\) where \(\alpha\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1+i\cot\theta =r(\cos\alpha +i \sin\alpha) \nonumber\\ \implies &\bf 1=r\cos\alpha \nonumber\\ \bf\bf~and~&\bf \cot\theta=r\sin\alpha\ \tag{1} \end{align}  squaring and adding \begin{align} &\bf r^{2}=\cosec^{2}\theta\nonumber\\ \implies &\bf r=- \cosec\theta\tag{2}\\ \end{align}  Since \(\bf \cosec\theta~\text{\textless}~0\) when \(\bf \frac{3\pi}{2}~ \text{\textless}~\theta~\text{\textless} ~2\pi \). Now putting (2) in (1), we get \( \\ \) \begin{align} &\bf \cos\alpha=-\sin\theta\nonumber\\ \bf~and~&\bf \sin\alpha=-\cos\theta\nonumber\\ \end{align}  then \begin{align} &\bf \alpha=\tan^{-1}{\left|\frac{-\cos\theta}{-\sin\theta}\right|} – \pi\nonumber \end{align}

Since if \(\bf \sin\alpha=a~and~\cos\alpha=b \) and
1. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right| \)
2. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\ge0\) then \(\bf \alpha=\pi-\tan^{-1}\left|\frac{b}{a}\right|\)
3. \(\bf \cos\alpha\le0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=\tan^{-1}\left|\frac{b}{a}\right|-\pi\)
4. \(\bf \cos\alpha\ge0 \) and \(\bf \sin\alpha\le0\) then \(\bf \alpha=-\tan^{-1}\left|\frac{b}{a}\right|\)

\begin{align} \implies &\bf \alpha= \tan^{-1}{\left[\cot\theta\right]}- \pi \nonumber\\ \implies &\bf \alpha= \tan^{-1}{\left[\tan\left(\frac{\pi}{2}-\theta\right)\right]}- \pi \nonumber\\ \implies &\bf \alpha= \frac{\pi}{2}-\theta- \pi \nonumber\\ \implies &\bf \alpha= -\left(\frac{\pi}{2}+\theta \right)\nonumber \end{align}

  Therefore the modulus amplitude form of \(\bf z=1+i\cot\theta\) is \( \bf 1+i\cot\theta=\cos\left(\frac{\pi}{2}+\theta \right) -i\sin\left(\frac{\pi}{2}+\theta \right) \) where \(\bf \frac{3\pi}{2}~ \text{\textless}~\theta~\text{\textless} ~2\pi \) .