Question

Convert the following into polar form!
$\sf z=\dfrac{i-1}{\cos\bigg(\dfrac{\pi}{3}\bigg)+i\;\sin\bigg(\dfrac{\pi}{3}\bigg)}$

Answer 1

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

Given Complex number is

$\rm :\longmapsto\:\sf z=\dfrac{i-1}{\cos\bigg(\dfrac{\pi}{3}\bigg)+i\;\sin\bigg(\dfrac{\pi}{3}\bigg)}$

We know,

$\boxed{ \tt{ \: cos\bigg[\dfrac{\pi}{3} \bigg] = \frac{1}{2}}}$

and

$\boxed{ \tt{ \: sin\bigg[\dfrac{\pi}{3} \bigg] = \frac{ \sqrt{3} }{2}}}$

So, above complex number can be rewritten as

$\rm :\longmapsto\:\sf z=\dfrac{i-1}{\dfrac{1}{2}+i\;\dfrac{ \sqrt{3} }{2} }$

$\rm \:  =  \:\dfrac{2(i - 1)}{1 + i \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \:  =  \:\dfrac{2(i - 1)}{1 + i \sqrt{3} } \times \dfrac{1 - i \sqrt{3} }{1 - i \sqrt{3} }$

$\rm \:  =  \:\dfrac{2(i - \sqrt{3} {i}^{2} - 1 + i\sqrt{3})}{ {1}^{2} - {( i\sqrt{3} )}^{2} }$

$\rm \:  =  \:\dfrac{2(i + \sqrt{3} - 1 + i\sqrt{3})}{ 1 - 3i^{2} }$

$\rm \:  =  \:\dfrac{2\bigg(\sqrt{3} - 1 + i(\sqrt{3} + 1)\bigg)}{ 1 + 3 }$

$\rm \:  =  \:\dfrac{2\bigg(\sqrt{3} - 1 + i(\sqrt{3} + 1)\bigg)}{4}$

$\rm \:  =  \:\dfrac{ \sqrt{3} - 1 + i( \sqrt{3} + 1) }{2}$

$\bf\implies \:z = \dfrac{ \sqrt{3} - 1 }{2} + i \: \dfrac{ \sqrt{3} + 1}{2}$

To represent z in polar form

Let we assume that,

$\rm :\longmapsto\:z = \dfrac{ \sqrt{3} - 1 }{2} + i \: \dfrac{ \sqrt{3} + 1}{2} = r(cosx + isinx) - - (1)$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ \sqrt{3} - 1 }{2} + i \: \dfrac{ \sqrt{3} + 1}{2} = rcosx + irsinx$

On comparing real and Imaginary parts, we get

$\rm :\longmapsto\:rcosx = \dfrac{ \sqrt{3} - 1}{2} - - - (2)$

and

$\rm :\longmapsto\:rsinx = \dfrac{ \sqrt{3} + 1}{2} - - - (3)$

On squaring equation (2) and (3) and adding we get,

$\rm :\longmapsto\: {r}^{2} {cos}^{2}x + {r}^{2} {sin}^{2}x = {\bigg[\dfrac{ \sqrt{3} - 1}{2} \bigg]}^{2} + {\bigg[\dfrac{ \sqrt{3} + 1}{2} \bigg]}^{2}$

$\rm :\longmapsto\: {r}^{2}({cos}^{2}x + {sin}^{2}x) = \dfrac{2( {( \sqrt{3}) }^{2} + {1}^{2}) }{4}$

$\red{\bigg \{ \because {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) \bigg \}}$

$\rm :\longmapsto\: {r}^{2} = \dfrac{3 + 1}{2}$

$\rm :\longmapsto\: {r}^{2} = \dfrac{4}{2}$

$\rm :\longmapsto\: {r}^{2} =2$

$\bf\implies \:r = \sqrt{2}$

On substituting, value of r in equation (2) and (3), we get

$\rm :\longmapsto\:cosx = \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} } - - - (4)$

and

$\rm :\longmapsto\:sinx = \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} } - - - (5)$

On dividing equation (5) by (4), we get

$\rm :\longmapsto\:\dfrac{sinx}{cosx} = \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1}$

$\rm :\longmapsto\:tanx = \dfrac{1 + \dfrac{1}{ \sqrt{3} } }{1 - \dfrac{1}{ \sqrt{3} } }$

$\rm :\longmapsto\:tanx = \dfrac{1 + \dfrac{1}{ \sqrt{3} } }{1 - \dfrac{1}{ \sqrt{3}} \times 1}$

$\rm :\longmapsto\:tanx = \dfrac{tan\bigg[\dfrac{\pi}{4} \bigg] + tan\bigg[\dfrac{\pi}{6} \bigg]}{1 - tan\bigg[\dfrac{\pi}{4} \bigg] \times tan\bigg[\dfrac{\pi}{6} \bigg]}$

$\rm :\longmapsto\:tanx = tan\bigg[\dfrac{\pi}{4} + \dfrac{\pi}{6} \bigg]$

$\rm :\longmapsto\:tanx = tan\bigg[\dfrac{3\pi + 2\pi}{12} \bigg]$

$\rm :\longmapsto\:tanx = tan\bigg[\dfrac{5\pi}{12} \bigg]$

$\bf\implies \:x = \dfrac{5\pi}{12}$

Hence, On substituting the values of x and r in equation (1), we get

$\rm :\longmapsto\:\dfrac{ \sqrt{3} - 1}{2} + i\dfrac{ \sqrt{3} + 1}{2} = \sqrt{2}\bigg(cos\bigg[\dfrac{5\pi}{12} \bigg] + isin\bigg[\dfrac{5\pi}{12} \bigg] \bigg)$

Hence,

The polar form of z is

$\boxed{ \tt{ \: \dfrac{i-1}{\cos\bigg(\dfrac{\pi}{3}\bigg)+i\;\sin\bigg(\dfrac{\pi}{3}\bigg)} = \sqrt{2}\bigg(cos\bigg[\dfrac{5\pi}{12} \bigg] + isin\bigg[\dfrac{5\pi}{12} \bigg] \bigg)}}$

Answer 2

Solution :-

We are given the complex number,

${ \implies z=\dfrac{i-1}{\cos\left(\dfrac{\pi}{3}\right)+i\;\sin\left(\dfrac{\pi}{3}\right)}}$

We are asked to represent the given complex number in its polar form.

For the polar representation of complex number, firstly we have to find the modulus i.e. the magnitude and the argument of the complex number. Let's firstly simply the given complex number to find the required result.

${ \implies z=\dfrac{i-1}{\cos\left(\dfrac{\pi}{3}\right)+i\;\sin\left(\dfrac{\pi}{3}\right)}}$

Substitute the values of,

• $\boxed{ \rm\cos \left( \dfrac \pi 3 \right ) = \dfrac{1}{2} }$

• $\boxed{ \rm\sin\left( \dfrac \pi 3 \right ) = \dfrac{ \sqrt{3} }{2} }$

We get,

${ \implies z=\dfrac{i-1}{ \dfrac{1}{2} + \dfrac{\sqrt{3}i}{2} }}$

${ \implies z=\dfrac{i-1}{\dfrac{1 + \sqrt{3}i}{2} }}$

${ \implies z=\dfrac{2(i-1)}{1 + \sqrt{3}i }}$

Now, rationalising the denominator to eliminate iota terms.

$\implies z=\dfrac{2(i-1)}{1 + \sqrt{3}i } \times \dfrac{1 - \sqrt{3} i}{1 - \sqrt{3} i}$

$\implies z=\dfrac{2i - 2}{1 + \sqrt{3}i } \times \dfrac{1 - \sqrt{3} i}{1 - \sqrt{3} i}$

$\implies z= \dfrac{(2i - 2)(1 - \sqrt{3} i)}{(1 + \sqrt{3}i)(1 - \sqrt{3} i)}$

${ \implies z=\dfrac{2i(1 - \sqrt{3} i) - 2(1 - \sqrt{3}i) }{1 - ( \sqrt{3} i)^{2} } }$

${ \implies z=\dfrac{2i - 2 \sqrt{3} {i}^{2} - 2 + 2 \sqrt{3}i}{1 -3 i^{2} } }$

${ \implies z=\dfrac{2i - 2 \sqrt{3} ( - 1)- 2 + 2 \sqrt{3}i}{1 -3 ( - 1)} \qquad( \because {i}^{2} = - 1)}$

${ \implies z=\dfrac{2i + 2 \sqrt{3}- 2 +2 \sqrt{3}i}{1 + 3}}$

${ \implies z=\dfrac{ (2 \sqrt{3}- 2 )+ 2i + 2\sqrt{3}i}{4}}$

${ \implies z=\dfrac{ 2( \sqrt{3}- 1 )+ 2(i + \sqrt{3})i}{4}}$

${ \implies z=\dfrac{ ( \sqrt{3}- 1 )+ (\sqrt{3} + 1)i}{2}}$

${ \implies z=\dfrac{( \sqrt{3}- 1 )}{2}+ \dfrac{(\sqrt{3} + 1)i}{2}}$

This is the simplified form of complex number.

Now let's find the modulus of the complex number.

$\leadsto|z|= \rm\sqrt{Re(z)^2+Im(z)^2}$

${\leadsto|z|= \rm\sqrt{ \bigg(\dfrac{ \sqrt{3}- 1 }{2} \bigg)^2+ \bigg(\dfrac{ \sqrt{3} + 1 }{2} \bigg)^2}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{ \sqrt{3}^{2} + 1 - 2 \sqrt{3} }{2}+ \dfrac{ \sqrt{3}^{2} + 1 + 2 \sqrt{3} }{2}}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{ 3 + 1 - 2 \sqrt{3} }{2}+ \dfrac{ 3+ 1 + 2 \sqrt{3} }{2}}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{ 4 - 2 \sqrt{3} }{2}+ \dfrac{4 + 2 \sqrt{3} }{2}}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{4 {}^{2} - (2 \sqrt{3} {)}^{2} }{2}}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{16-12 }{2}}}$

${\leadsto|z|= \rm\sqrt{ \dfrac{4 }{2}}}$

$\boxed{\leadsto|z|= \rm\sqrt{ 2}}$

Now let's find the argument of the complex number

The given complex number will lie in 1st quadrant because both real and imaginary part of complex number is positive.

Argument in 1st quadrant is given by,

$\dashrightarrow \theta = \rm\tan^{ - } \left | \dfrac{ Im (z)}{Re(z)} \right |$

$\dashrightarrow \theta = \tan^{ - } \left | \dfrac{\dfrac{(\sqrt{3} + 1)i}{2}}{\dfrac{(\sqrt{3} - 1)i}{2}} \right |$

$\dashrightarrow \theta = \tan^{ - } \left | \dfrac{(\sqrt{3} + 1)}{(\sqrt{3} - 1)} \right |$

${ \dashrightarrow \theta = \tan^{ - } \left | \dfrac{(\sqrt{3} + 1)}{(\sqrt{3} - 1)} \times \dfrac{( \sqrt{3} + 1)}{ (\sqrt{3 } + 1) } \right | }$

${ \dashrightarrow \theta = \tan^{ - } \left | \dfrac{(\sqrt{3} )^{2} + 1 + 2 \sqrt{3} }{(\sqrt{3})^{2} - 1} \right | }$

${ \dashrightarrow\theta = \tan^{ - } \left | \dfrac{3 + 1 + 2 \sqrt{3} }{3 - 1} \right | }$

${\dashrightarrow \theta = \tan^{ - } \left | \dfrac{4+ 2 \sqrt{3} }{2} \right | }$

${ \dashrightarrow \theta = \tan^{ - } \left |2 + \sqrt{3 } \right | }$

$\boxed{ \dashrightarrow \theta = \dfrac{5\pi}{12} }$

Polar form of a complex number is given by,

$\longrightarrow |z| \bigg( \cos \theta +i \sin \theta \bigg)$

${ \longrightarrow \underline{ \underline{ \sqrt{2} \bigg( \cos \left[ \dfrac{5\pi}{12} \right] +i \sin \left[ \dfrac{5\pi}{12} \right] \bigg)}}}$

This is the required polar form of given complex number.