Question

2. The polar form of i
$the \: polar \: form \: of \: i {75} \: is$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: {i}^{75}$

Let assume that

$\rm :\longmapsto\: z = {i}^{75}$

can be rewritten as

$\rm :\longmapsto\: z = {i}^{74} \times i$

$\rm :\longmapsto\: z = {( {i}^{2}) }^{37} \times i$

$\rm :\longmapsto\: z = {( - 1)}^{37} \times i$

$\rm :\longmapsto\: z = - 1 \times i$

$\rm :\longmapsto\: z = - i$

is the standard form and now we have to represent this in standard form.

So, the above complex number have to be represent in the form

$\boxed{\tt{ z \: = \: r \: (cos \theta \: + \: i \: sin \theta \: }}$

So, above complex number can be rewritten as

$\rm :\longmapsto\: z = 0 - i$

$\rm :\longmapsto\: z = cos\bigg(\dfrac{\pi}{2} \bigg) - isin\bigg(\dfrac{\pi}{2} \bigg)$

We know,

$\boxed{\tt{ cos( - x) = cosx}} \: \: and \: \: \boxed{\tt{ sin( - x) - sinx}}$

So, using this, we get

$\rm :\longmapsto\: z = cos\bigg( - \dfrac{\pi}{2} \bigg) + i \: sin\bigg( - \dfrac{\pi}{2} \bigg)$

can be rewritten as

$\rm :\longmapsto\: z = 1\bigg[cos\bigg( - \dfrac{\pi}{2} \bigg) + i \: sin\bigg( - \dfrac{\pi}{2} \bigg)\bigg]$

which is the required polar form having

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{ |z| = 1} \\ \\ &\sf{arg(z) = - \dfrac{\pi}{2} } \end{cases}\end{gathered}\end{gathered}$

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More to Know

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: {i}^{75}$

Let assume that

$\rm :\longmapsto\: z = {i}^{75}$

can be rewritten as

$\rm :\longmapsto\: z = {i}^{74} \times i$

$\rm :\longmapsto\: z = {( {i}^{2}) }^{37} \times i$

$\rm :\longmapsto\: z = {( - 1)}^{37} \times i$

$\rm :\longmapsto\: z = - 1 \times i$

$\rm :\longmapsto\: z = - i$

is the standard form and now we have to represent this in standard form.

So, the above complex number have to be represent in the form

$\boxed{\tt{ z \: = \: r \: (cos \theta \: + \: i \: sin \theta \: }}$

So, above complex number can be rewritten as

$\rm :\longmapsto\: z = 0 - i$

$\rm :\longmapsto\: z = cos\bigg(\dfrac{\pi}{2} \bigg) - isin\bigg(\dfrac{\pi}{2} \bigg)$

We know,

$\boxed{\tt{ cos( - x) = cosx}} \: \: and \: \: \boxed{\tt{ sin( - x) - sinx}}$

So, using this, we get

$\rm :\longmapsto\: z = cos\bigg( - \dfrac{\pi}{2} \bigg) + i \: sin\bigg( - \dfrac{\pi}{2} \bigg)$

can be rewritten as

$\rm :\longmapsto\: z = 1\bigg[cos\bigg( - \dfrac{\pi}{2} \bigg) + i \: sin\bigg( - \dfrac{\pi}{2} \bigg)\bigg]$

which is the required polar form having

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{ |z| = 1} \\ \\ &\sf{arg(z) = - \dfrac{\pi}{2} } \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$