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 $$\begin{array}{rl}{\displaystyle }& {\displaystyle \mathbf{a}+\mathbf{i}\mathbf{b}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{a}=\mathbf{r}\cos \mathbf{\theta }\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{b}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}={\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\sqrt{{\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \mathbf{a}=\sqrt{{\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{b}=\sqrt{{\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\frac{\mathbf{a}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\frac{\mathbf{b}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}+{\mathbf{b}}^{\mathbf{2}}}}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }={\tan }^{-\mathbf{1}}\mid \frac{\mathbf{b}}{\mathbf{a}}\mid }\end{array}$$$$\begin{align} &\bf a+ib=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf a=r\cos\theta~and~ b=r\sin\theta\ \\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=a^{2}+b^{2}\nonumber\\ \implies &\bf r=\sqrt{a^{2}+b^{2}}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf a=\sqrt{a^{2}+b^{2}}\cos\theta\nonumber\\ \bf~and~&\bf b=\sqrt{a^{2}+b^{2}}\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{a}{\sqrt{a^{2}+b^{2}}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{b}{\sqrt{a^{2}+b^{2}}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{b}{a}\right|}\nonumber \end{align}$$

 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{array}{rl}{\displaystyle }& {\displaystyle \mathbf{0}+\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{0}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{1}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\mathbf{1}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \mathbf{0}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{1}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\mathbf{0}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\mathbf{1}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }={\tan }^{-\mathbf{1}}\mid \frac{\mathbf{1}}{\mathbf{0}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\frac{\mathbf{\pi }}{\mathbf{2}}}\end{array}$$$$\begin{align} &\bf 0+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 0=r\cos\theta \nonumber\\ \bf\bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1 \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 0=\cos\theta\nonumber\\ \bf~and~&\bf 1=\sin\theta\nonumber\\ \implies &\bf \cos\theta=0\nonumber\\ \bf~and~&\bf \sin\theta=1\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{1}{0}\right|}\nonumber\\ \implies &\bf \theta=\frac{\pi}{2} \nonumber \end{align}$$   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{array}{rl}{\displaystyle }& {\displaystyle \mathbf{1}+\mathbf{0.}\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{0}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{1}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\mathbf{1}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \mathbf{1}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{0}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\mathbf{1}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\mathbf{0}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }={\tan }^{-\mathbf{1}}\mid \frac{\mathbf{0}}{\mathbf{1}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\mathbf{0}}\end{array}$$$$\begin{align} &\bf 1+0.i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf 0=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 1=\cos\theta\nonumber\\ \bf~and~&\bf 0=\sin\theta\nonumber\\ \implies &\bf \cos\theta=1\nonumber\\ \bf~and~&\bf \sin\theta=0\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{0}{1}\right|}\nonumber\\ \implies &\bf \theta=0 \nonumber \end{align}$$   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{array}{rl}{\displaystyle }& {\displaystyle \mathbf{0}-\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{0}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{1}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\mathbf{1}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \mathbf{0}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\mathbf{1}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\mathbf{0}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=-\mathbf{1}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }=-{\tan }^{-\mathbf{1}}\mid \frac{-\mathbf{1}}{\mathbf{0}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=-\frac{\mathbf{\pi }}{\mathbf{2}}}\end{array}$$$$\begin{align} &\bf 0-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 0=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf 0=\cos\theta\nonumber\\ \bf~and~&\bf -1=\sin\theta\nonumber\\ \implies &\bf \cos\theta=0\nonumber\\ \bf~and~&\bf \sin\theta=-1\nonumber\\ \bf~then~&\bf \theta=-\tan^{-1}{\left|\frac{-1}{0}\right|}\nonumber\\ \implies &\bf \theta=-\frac{\pi}{2} \nonumber \end{align}$$   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{array}{rl}{\displaystyle -\mathbf{1}+\mathbf{0.}\mathbf{i}}& {\displaystyle =\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle -\mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{0}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{1}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\mathbf{1}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle -\mathbf{1}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{0}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=-\mathbf{1}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\mathbf{0}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }=\mathbf{\pi }-{\tan }^{-\mathbf{1}}\mid \frac{\mathbf{0}}{-\mathbf{1}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\mathbf{\pi }}\end{array}$$$$\begin{align} \bf -1+0.i&\bf =r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf 0=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=1\nonumber\\ \implies &\bf r=1\tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -1=\cos\theta\nonumber\\ \bf~and~&\bf 0=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-1\nonumber\\ \bf~and~&\bf \sin\theta=0\nonumber\\ \bf~then~&\bf \theta=\pi-\tan^{-1}{\left|\frac{0}{-1}\right|}\nonumber\\ \implies &\bf \theta=\pi \nonumber \end{align}$$   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 $$\begin{array}{rl}{\displaystyle }& {\displaystyle \mathbf{1}+\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{2}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\sqrt{\mathbf{2}}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }={\tan }^{-\mathbf{1}}\mid \frac{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\frac{\mathbf{\pi }}{\mathbf{4}}}\end{array}$$$$\begin{align} &\bf 1+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf \frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf \frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=\frac{\pi}{4} \nonumber \end{align}$$   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 $$\begin{array}{rl}{\displaystyle }& {\displaystyle \mathbf{1}-\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{2}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\sqrt{\mathbf{2}}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle \frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }=-{\tan }^{-\mathbf{1}}\mid \frac{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}{-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=-\frac{\mathbf{\pi }}{\mathbf{4}}}\end{array}$$$$\begin{align} &\bf 1-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf 1=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf \frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf -\frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=-\tan^{-1}{\left|\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=-\frac{\pi}{4} \nonumber \end{align}$$   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 $$\begin{array}{rl}{\displaystyle }& {\displaystyle -\mathbf{1}+\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle -\mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{2}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\sqrt{\mathbf{2}}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle -\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }=\mathbf{\pi }-{\tan }^{-\mathbf{1}}\mid \frac{-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\mid }\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\mathbf{\pi }-\frac{\mathbf{\pi }}{\mathbf{4}}=\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}}\end{array}$$$$\begin{align} &\bf -1+i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf 1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -\frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf \frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\pi-\tan^{-1}{\left|\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right|}\nonumber\\ \implies &\bf \theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4} \nonumber \end{align}$$   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 $$\begin{array}{rl}{\displaystyle }& {\displaystyle -\mathbf{1}-\mathbf{i}=\mathbf{r}(\cos \mathbf{\theta }+\mathbf{i}\sin \mathbf{\theta })}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle -\mathbf{1}=\mathbf{r}\cos \mathbf{\theta }}\\ & {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\mathbf{1}=\mathbf{r}\sin \mathbf{\theta }}& & \\ {\displaystyle }& {\displaystyle \mathbf{s}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{d}\mathbf{d}\mathbf{i}\mathbf{n}\mathbf{g}}\\ {\displaystyle }& {\displaystyle {\mathbf{r}}^{\mathbf{2}}=\mathbf{2}}\\ & {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{r}=\sqrt{\mathbf{2}}}& & \\ {\displaystyle }& {\displaystyle \mathbf{p}\mathbf{u}\mathbf{t}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}(\mathbf{2})\mathbf{i}\mathbf{n}(\mathbf{1}),}\\ {\displaystyle }& {\displaystyle -\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\cos \mathbf{\theta }}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle -\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}=\sin \mathbf{\theta }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \cos \mathbf{\theta }=-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{a}\mathbf{n}\mathbf{d}}& {\displaystyle \sin \mathbf{\theta }=-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\\ {\displaystyle \mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}}& {\displaystyle \mathbf{\theta }={\tan }^{-\mathbf{1}}\mid \frac{-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}{-\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\mid -\mathbf{\pi }}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle \mathbf{\theta }=\frac{\mathbf{\pi }}{\mathbf{4}}-\mathbf{\pi }=-\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}}\end{array}$$$$\begin{align} &\bf -1-i=r(\cos\theta +i \sin\theta) \nonumber\\ \implies &\bf -1=r\cos\theta \nonumber\\ \bf~and~&\bf -1=r\sin\theta\ \tag{1}\\ &\bf squaring~~and~~adding\nonumber\\ &\bf r^{2}=2\nonumber\\ \implies &\bf r=\sqrt{2} \tag{2}\\ &\bf putting~~(2)~~in~~(1),\nonumber\\ &\bf -\frac{1}{\sqrt{2}}=\cos\theta\nonumber\\ \bf~and~&\bf -\frac{1}{\sqrt{2}}=\sin\theta\nonumber\\ \implies &\bf \cos\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~and~&\bf \sin\theta=-\frac{1}{\sqrt{2}}\nonumber\\ \bf~then~&\bf \theta=\tan^{-1}{\left|\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right|}-\pi\nonumber\\ \implies &\bf \theta=\frac{\pi}{4}-\pi=-\frac{3\pi}{4} \nonumber \end{align}$$   There the modulus amplitude form of $\bf -1-i$ is $\bf -1-i=\sqrt{2}\left(\cos\frac{3\pi}{4} -i\sin\frac{3\pi}{4}\right)$