Question

Convert into polar form \[z = \dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}\].

Answer 1

Hint: First we will first use the value of \[\pi \] in the given equation to simplify the given equation by rationalizing it. Then compare the real and imaginary values of the obtained equation the polar form of \[z = r\cos \theta + i\sin \theta \] to find the required values.

Complete step by step solution: We are given that \[z = \dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}\].

Using the value of \[\pi \] in the above equation, we get
\[
   \Rightarrow z = \dfrac{{i - 1}}{{\cos \left( {\dfrac{{180}}{3}} \right) + i\sin \left( {\dfrac{{180}}{3}} \right)}} \\
   \Rightarrow z = \dfrac{{i - 1}}{{\cos 60^\circ + i\sin 60^\circ }} \\
   \Rightarrow z = \dfrac{{i - 1}}{{\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}}} \\
   \Rightarrow z = \dfrac{{i - 1}}{{\dfrac{{1 + i\sqrt 3 }}{2}}} \\
   \Rightarrow z = \dfrac{{2\left( {i - 1} \right)}}{{1 + \sqrt 3 i}} \\
 \]

Rationalizing the above equation by multiplying the numerator and denominator with \[1 - \sqrt 3 i\], we get
\[
   \Rightarrow z = \dfrac{{2\left( {i - 1} \right) \times \left( {1 - \sqrt 3 i} \right)}}{{\left( {1 + \sqrt 3 i} \right) \times \left( {1 - \sqrt 3 i} \right)}} \\
   \Rightarrow z = \dfrac{{2\left( {i - 1} \right) \times \left( {1 - \sqrt 3 i} \right)}}{{{1^2} - {{\left( {\sqrt 3 i} \right)}^2}}} \\
   \Rightarrow z = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i - \sqrt 3 i\left( {{i^2}} \right)} \right]}}{{{1^2} - 3{i^2}}} \\
 \]

Putting \[{i^2} = - 1\] in the above equation, we get
\[
   \Rightarrow z = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i - \sqrt 3 \left( { - 1} \right)} \right]}}{{1 - 3\left( { - 1} \right)}} \\
   \Rightarrow z = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i + \sqrt 3 } \right]}}{{1 + 3}} \\
   \Rightarrow z = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i + \sqrt 3 } \right]}}{4} \\
   \Rightarrow z = \dfrac{{\left[ { - 1 + i + \sqrt 3 i + \sqrt 3 } \right]}}{2} \\
   \Rightarrow z = \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2}{\text{ .......eq.(1)}} \\
 \]

Let us assume that the polar form be
\[z = r\left( {\cos \theta + i\sin \theta } \right){\text{ ......eq.(2)}}\]

From equation (1) and equation (2), we get
\[
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} = r\left( {\cos \theta + i\sin \theta } \right) \\
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} = r\cos \theta + ir\sin \theta {\text{ ......eq.(3)}} \\
 \]

Comparing the real parts in the above equation (3), we get
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{2} = r\cos \theta \]

Squaring the above equation on both sides, we get
\[
   \Rightarrow {\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)^2} = {\left( {r\cos \theta } \right)^2} \\
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{2^2}}} = {r^2}{\cos ^2}\theta \\
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 } \right)}^2} + 1 - 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta \\
   \Rightarrow \dfrac{{3 + 1 - 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta \\
   \Rightarrow \dfrac{{4 - 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta {\text{ ......eq.(4)}} \\
 \]

Comparing the imaginary parts in the equation (3), we get
\[ \Rightarrow \dfrac{{\sqrt 3 + 1}}{2} = r\sin \theta \]

Squaring the above equation on both sides, we get
\[
   \Rightarrow {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2} = {\left( {r\sin \theta } \right)^2} \\
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{{2^2}}} = {r^2}{\sin ^2}\theta \\
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 } \right)}^2} + 1 + 2\sqrt 3 }}{4} = {r^2}{\sin ^2}\theta \\
   \Rightarrow \dfrac{{3 + 1 + 2\sqrt 3 }}{4} = {r^2}{\sin ^2}\theta \\
   \Rightarrow \dfrac{{4 + 2\sqrt 3 }}{4} = {r^2}{\sin ^2}\theta {\text{ .......eq.(5)}} \\
 \]

Adding the equation (3) and (4), we get
\[
   \Rightarrow \dfrac{{4 - 2\sqrt 3 }}{4} + \dfrac{{4 + 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta \\
   \Rightarrow \dfrac{{4 - 2\sqrt 3 + 4 + 2\sqrt 3 }}{4} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
   \Rightarrow \dfrac{8}{4} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
   \Rightarrow 2 = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
 \]

Using \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] in the above equation, we get
\[
   \Rightarrow 2 = {r^2}\left( 1 \right) \\
   \Rightarrow 2 = {r^2} \\
 \]

Taking square root on both sides in the above equation, we get
\[
   \Rightarrow \sqrt 2 = r \\
   \Rightarrow r = \sqrt 2 \\
 \]

Substituting the value of \[r\] in the equation (3), we get
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} = \sqrt 2 \cos \theta + \sqrt 2 i\sin \theta \]

Dividing the above equation by \[\sqrt 2 \] on both sides, we get
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + i\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \cos \theta + i\sin \theta {\text{ .......eq.(6)}}\]

Comparing the real parts in the above equation (6), we get
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} = \cos \theta \]

Comparing the imaginary parts in the equation (6), we get
\[ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \sin \theta \]

Thus, the polar form is \[z = \sqrt 2 \left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right)\].

Note: In solving these types of questions, students should know the standard form of the polar form of the equation is \[z = r\cos \theta + i\sin \theta \]. Whenever we have asked to convert the complex number to polar coordinates, we try to convert in \[z = r\cos \theta + i\sin \theta \] form.