Modulus-Amplitude form

Modulus-Amplitude form #

  Definition: Let \(\bf z=a+ib\) be a non-complex number and \(\bf |z|=r\). Then \(\bf z\) can represented as \(\bf z=r(\cos\theta +i \sin\theta)\). And this form of \(\bf z\) is said to be the \(\fcolorbox{red}{white}{\bf modulus-amplitude}\) or \(\fcolorbox{red}{white}{\bf polar}\) form of \(\bf z\). The \(\fcolorbox{red}{white}{\bf \( \theta\)}\) is said to be a amplitude or argument of \(\bf z\) and denoted by \(\fcolorbox{red}{white}{\bf \( amp~z=\theta \)}\). If \( \theta\) satistifes the condition \( -\pi\text{\textless}\theta\le \pi\) then \( \theta\) is said to the \(\fcolorbox{red}{white}{\bf principal amplitude}\) of \( z\). \( \\ \)

Method of finding Principal Amplitude #

 Problem:  Find the principal amplitude of the complex number \(\bf z=a+ib\).\( \\ \)  Solution:  Given that \(\bf z=a+ib\).\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\).\( \\ \)   Then \begin{align} &\bf a+ib=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf a=r\cos\theta~and~ b=r\sin\theta\ \\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=a^{2}+b^{2}\nonumber\\ \implies &\bf r=\sqrt{a^{2}+b^{2}}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf a=\sqrt{a^{2}+b^{2}}\cos\theta\nonumber\\ \bf~and~&\bf b=\sqrt{a^{2}+b^{2}}\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{a}{\sqrt{a^{2}+b^{2}}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{b}{\sqrt{a^{2}+b^{2}}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{b}{a}\right|}\nonumber \end{align}

 Let \(\bf \alpha\) be the principal amplitude of \(\bf z\). If\( \\ \)  (1)  \(\bf \cos\theta\ge0 \) and \(\bf \sin\theta\ge0\) then \(\fcolorbox{blue}{white}{\bf \(\bf \alpha=\theta\)}\) \( \\ \)  (2)  \(\bf \cos\theta\le0 \) and \(\bf \sin\theta\ge0\) then \(\fcolorbox{blue}{white}{\bf \(\bf \alpha=\pi-\theta\)}\) \( \\ \)  (3)  \(\bf \cos\theta\le0 \) and \(\bf \sin\theta\le0\) then \(\fcolorbox{blue}{white}{\bf \(\bf \alpha=\theta-\pi\)}\) \( \\ \)  (4)  \(\bf \cos\theta\ge0 \) and \(\bf \sin\theta\le0\) then \(\fcolorbox{blue}{white}{\bf \(\bf \alpha=-\theta\)}\) \( \\ \)

Examples #

Modulus-Amplitude form of \(\bf i\) #

  Problem-1: Find the modulus amplitude form of \(\bf i\).\( \\ \) Solution:  Given that \(\bf z=0+i\)\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 0+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 0=r\cos\theta \nonumber\\ \bf\bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1 \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 0=\cos\theta\nonumber\\ \bf~and~&\bf 1=\sin\theta\nonumber\\ \implies &\bf \cos\theta=0\nonumber\\ \bf~and~&\bf \sin\theta=1\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{1}{0}\right|}\nonumber\\ \implies &\bf \theta=\frac{\pi}{2} \nonumber \end{align}   There the modulus amplitude form of \(\bf i\) is \( \bf i=\cos\frac{\pi}{2} +i\sin\frac{\pi}{2} \)

Modulus-Amplitude form of 1 #

  Problem-2: Find the modulus amplitude form of 1.\( \\ \) Solution:  Given that \(\bf z=1+0.i\)\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1+0.i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf 0=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 1=\cos\theta\nonumber\\ \bf~and~&\bf 0=\sin\theta\nonumber\\ \implies &\bf \cos\theta=1\nonumber\\ \bf~and~&\bf \sin\theta=0\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{0}{1}\right|}\nonumber\\ \implies &\bf \theta=0 \nonumber \end{align}   There the modulus amplitude form of \(\bf 1\) is \(\bf 1=\cos0 +i \sin0 \)

Modulus-Amplitude form of \(\bf -i \) #

  Problem-3: Find the modulus amplitude form of \(\bf -i \).\( \\ \) Solution:  Given that \(\bf z=0-i\)\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 0-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 0=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 0=\cos\theta\nonumber\\ \bf~and~&\bf -1=\sin\theta\nonumber\\ \implies &\bf \cos\theta=0\nonumber\\ \bf~and~&\bf \sin\theta=-1\nonumber\\ \bf~then~&\bf \theta=-\tan^{-1}{\left|\frac{-1}{0}\right|}\nonumber\\ \implies &\bf \theta=-\frac{\pi}{2} \nonumber \end{align}   There the modulus amplitude form of \(\bf -i\) is \(\bf -i=\cos\frac{\pi}{2} -i \sin\frac{\pi}{2} \)

Modulus-Amplitude form of \(\bf -1 \) #

  Problem-4: Find the modulus amplitude form of \(\bf -1 \).\( \\ \) Solution:  Given that \(\bf z=-1+0.i\)\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} \bf -1+0.i&\bf =r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf 0=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -1=\cos\theta\nonumber\\ \bf~and~&\bf 0=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-1\nonumber\\ \bf~and~&\bf \sin\theta=0\nonumber\\ \bf~then~&\bf \theta=\pi-\tan^{-1}{\left|\frac{0}{-1}\right|}\nonumber\\ \implies &\bf \theta=\pi \nonumber \end{align}   There the modulus amplitude form of \(\bf -1\) is \(\bf -1=\cos\pi+i \sin\pi \)

Modulus-Amplitude form of \(\bf 1+i\) #

  Problem-5: Find the modulus amplitude form of \(\bf 1+i\).\( \\ \) Solution:  Given that \(\bf z=1+i\).\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf \frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf \frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=\frac{\pi}{4} \nonumber \end{align}   There the modulus amplitude form of \(\bf 1+i\) is \( \bf 1+i=\sqrt{2}\left(\cos\frac{\pi}{4} +i\sin\frac{\pi}{4}\right) \)

Modulus-Amplitude form of \(\bf 1-i\) #

  Problem-6: Find the modulus amplitude form of \(\bf 1-i\).\( \\ \) Solution:  Given that \(\bf z=1-i\).\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf 1-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf \frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf -\frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=-\tan^{-1}{\left|\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=-\frac{\pi}{4} \nonumber \end{align}   There the modulus amplitude form of \(\bf 1-i\) is \( \bf 1-i=\sqrt{2}\left(\cos\frac{\pi}{4} -i\sin\frac{\pi}{4}\right) \)

Modulus-Amplitude form of \(\bf -1+i\) #

  Problem-7: Find the modulus amplitude form of \(\bf -1+i\).\( \\ \) Solution:  Given that \(\bf z=-1+i\).\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf -1+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -\frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf \frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\pi-\tan^{-1}{\left|\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4} \nonumber \end{align}   There the modulus amplitude form of \(\bf -1+i\) is \( \bf -1+i=\sqrt{2}\left(\cos\frac{3\pi}{4} +i\sin\frac{3\pi}{4}\right) \)

Modulus-Amplitude form of \(\bf -1-i\) #

  Problem-8: Find the modulus amplitude form of \(\bf -1-i\).\( \\ \) Solution:  Given that \(\bf z=-1-i\).\( \\ \)   Let \(\bf z=r(\cos\theta +i \sin\theta)\) where \(\theta\) be the principal amplitude .\( \\ \)   Then \begin{align} &\bf -1-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -\frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf -\frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right|}-\pi\nonumber\\ \implies &\bf \theta=\frac{\pi}{4}-\pi=-\frac{3\pi}{4} \nonumber \end{align}   There the modulus amplitude form of \(\bf -1-i\) is \( \bf -1-i=\sqrt{2}\left(\cos\frac{3\pi}{4} -i\sin\frac{3\pi}{4}\right) \)