Question

Convert the following into polar form, \[\dfrac{1+7i}{{{\left( 2-i \right)}^{2}}}\] .

Answer 1

Hint: The given complex number is \[\dfrac{1+7i}{{{\left( 2-i \right)}^{2}}}\] . Expand the denominator of the given complex number using the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] . Now, substitute \[{{i}^{2}}\] by -1 and simplify it. Now, multiply by the term \[\left( 3+4i \right)\] in numerator and denominator of \[\dfrac{1+7i}{\left( 3-4i \right)}\] . Again, substitute \[{{i}^{2}}\] by -1 and simplify it. We know the standard form of a complex number in the polar form is of the form, \[r\left( \cos \theta +i\sin \theta \right)\] . Compare \[r\left( \cos \theta +i\sin \theta \right)\] and \[\left( -1+i \right)\] to get the value of \[r\cos \theta \] and \[r\sin \theta \] . Now, add after squaring and then, use the identity \[\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=1\] to get the value of \[r\] . Now, put the value of \[r\] in \[r\cos \theta =\dfrac{-1}{\sqrt{2}}\] and
\[r\sin \theta =\dfrac{1}{\sqrt{2}}\] to get the value of \[\theta \] . At last, put the value of \[\theta \] and \[r\] in \[r\left( \cos \theta +i\sin \theta \right)\] . Now, solve it further and get the polar form of the given complex number.

Complete step by step answer:

According to the question, it is given that
The complex number = \[\dfrac{1+7i}{{{\left( 2-i \right)}^{2}}}\] ……………………………………..(1)
The denominator of the given complex number = \[{{\left( 2-i \right)}^{2}}\] ……………………………………………(2)
We know the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] ……………………………………………(3)
Now, from equation (2) and equation (3), we get
\[{{\left( 2-i \right)}^{2}}={{\left( 2 \right)}^{2}}+{{\left( i \right)}^{2}}-2\left( 2 \right)\left( i \right)\] ………………………………………..(4)
We know that \[{{i}^{2}}=-1\] …………………………………..(5)
Now, on substituting \[{{i}^{2}}\] by -1 in equation (4), we get
\[{{\left( 2-i \right)}^{2}}=4-1-4\left( i \right)=3-4i\] ………………………………………….(6)
On replacing \[{{\left( 2-i \right)}^{2}}\] by \[\left( 3-4i \right)\] in equation (1), we get
The complex number = \[\dfrac{1+7i}{\left( 3-4i \right)}\] …………………………………………(7)
Now, on multiplying by the term \[\left( 3+4i \right)\] in numerator and denominator of equation (7), we get
The complex number = \[\dfrac{\left( 1+7i \right)\left( 3+4i \right)}{\left( 3-4i \right)\left( 3+4i \right)}=\dfrac{3+4i+21i+28{{i}^{2}}}{9-16{{i}^{2}}}\] ……………………………………..(8)
Substituting \[{{i}^{2}}\] by -1 in equation (8), we get
The complex number = \[\dfrac{3+25i+28\left( -1 \right)}{9-16\left( -1 \right)}=\dfrac{-25+25i}{9+16}=-1+i\] …………………………………………(9)
We know the standard form of a complex number in the polar form is of the form, \[r\left( \cos \theta +i\sin \theta \right)\] ……………………………………(10)
We have to convert the given complex number into polar form.
Now, on comparing equation (9) and equation (10), we get
\[\Rightarrow r\left( \cos \theta +i\sin \theta \right)=\left( -1+i \right)\]
\[\Rightarrow r\cos \theta +i\left( r\sin \theta \right)=-1+i\] ……………………………………(11)
In equation (11), since the LHS is equal to the RHS so, the real part of the LHS must be equal to the real part of the RHS. Similarly, the imaginary part of the LHS must be equal to the imaginary part of the RHS.
Therefore,
\[r\cos \theta =-1\] ………………………………………(12)
\[r\sin \theta =1\] …………………………………………(13)
Now, on squaring and adding equation (12) and equation (13), we get
\[\Rightarrow {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =1+1\]
\[\Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=2\] ……………………………………..(14)
We know the identity, \[\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=1\] …………………………………….(15)
Now, from equation (14) and equation (15), we get
\[\Rightarrow {{r}^{2}}=2\]
\[\Rightarrow r=\sqrt{2}\] …………………………………..(16)
On substituting \[r\] by \[\sqrt{2}\] in equation (12), we get
\[\Rightarrow \sqrt{2}\cos \theta =-1\]
\[\Rightarrow \cos \theta =\dfrac{-1}{\sqrt{2}}\] ………………………………………..(17)
Similarly, on substituting \[r\] by \[\sqrt{2}\] in equation (13), we get
\[\Rightarrow \sqrt{2}sin\theta =-1\]
\[\Rightarrow sin\theta =\dfrac{1}{\sqrt{2}}\] ………………………………………..(18)
We know the property that only in quadrant II, \[\cos \theta \] is negative and \[sin\theta \] is positive ……………………………………(19)
In equation (17) and equation (18), we can observe that \[\cos \theta \] is negative and \[sin\theta \] is positive.
Using the property shown in equation (19), we can say that \[\theta \] is lying in quadrant II.
We also know that \[\cos \dfrac{3\pi }{4}=\dfrac{-1}{\sqrt{2}}\] and \[\sin \dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}\] ……………………………………………(20)
 Now, from equation (17), and equation (20), we get
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \dfrac{3\pi }{4} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \cos \dfrac{3\pi }{4} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{3\pi }{4}\] ………………………………………….(21)
Similarly, from equation (18), and equation (20), we get
\[\begin{align}
  & \Rightarrow \sin \theta =\sin \dfrac{3\pi }{4} \\
 & \Rightarrow \theta ={{\sin }^{-1}}\left( \sin \dfrac{3\pi }{4} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{3\pi }{4}\] ………………………………………….(22)
From equation (21) and equation (22), we have the value of \[\theta \] .
Therefore, the value of \[\theta \] is \[\dfrac{3\pi }{4}\] , \[\theta =\dfrac{3\pi }{4}\] …………………………………(23)
Now, from equation (10), equation (16), and equation (23), we have
Complex number in polar form = \[\sqrt{2}\left( \cos \dfrac{3\pi }{4}+i\sin \dfrac{3\pi }{4} \right)\] .
Therefore, the polar form of the given complex number is \[\sqrt{2}\left( \cos \dfrac{3\pi }{4}+i\sin \dfrac{3\pi }{4} \right)\] .

Note:
In this question, one might try to convert the given complex number directly into polar form without simplifying it in the form of \[a+ib\]. This approach is very complex which may result in calculation mistakes and sometimes we can’t convert the given complex number directly into the polar form. Therefore, the best way to approach this type of question is to simplify the given complex number in the form of \[\left( a+ib \right)\] and then, compare it with the standard form of the polar form, \[r\left( \cos \theta +i\sin \theta \right)\] to get the value of \[r\] and \[\theta \] .