Question

The complex number $z = 1 + i$ represented by the point $P$ in argand plane and $OP$ is rotated by and an angle of $\dfrac{\pi }{2}$in counter clockwise direction then the resulting complex number is :
A) $\overline z $
B) $z$
C) $ - \overline z $
D) $\dfrac{{\overline z }}{{|z{|^2}}}$

Answer 1

Hint: A complex number of the form $z = a + ib$, can be represented in the form of $r{e^{i\theta }}$ which can further be written as $r(\cos \theta + i\sin \theta )$, where $r$ is the modulus of the complex number $z$ and $\theta $ is the angle made by the complex number with the real axis in an argand plane. If a complex number $z$ is rotated by an angle $\theta $ in the anticlockwise direction, then we get the answer to be $z{e^{i\theta }}$. Then, we should know the condition that if a complex number $z$ is rotated by an angle $\theta $ in the clockwise direction, then we get the answer to be $z{e^{ - i\theta }}$.

Complete step by step answer:
We are given a complex number $z = 1 + i$ represented by the point $P$ in argand plane and $OP$ is rotated by and an angle of $\dfrac{\pi }{2}$in counter clock wise direction.
Represent the given complex number in the form of $r{e^{i\theta }}$, where $r$is the modulus of the complex number and $\theta $ is the angle made by the complex number with the real axis in the argand plane.
For any complex number of the form $z = a + bi$, the angle made by the given complex number with the real axis, is given by $\theta = {\tan ^{ - 1}}\dfrac{b}{a}$.
For$z = 1 + i$, determine the value of $\theta $,
$\theta = {\tan ^{ - 1}}\dfrac{1}{1}$
$\theta = {\tan ^{ - 1}}1$
$
  \theta = {45^ \circ } \\
  \theta = \dfrac{\pi }{4} \\
 $
So for the complex number $z = 1 + i$ represented by the point $P$ , the angle line $OP$ makes with the real axis in an argand plane is $\dfrac{\pi }{4}$.
Now, determine the length of $OP$, which is evaluated by taking the modulus of the given complex number.
For a complex number $z = a + bi$, the modulus of this complex number is given by $|z| = \sqrt {{a^2} + {b^2}} $
For $z = 1 + i$,
$|z| = \sqrt {{1^2} + {1^2}} = \sqrt {1 + 1} = \sqrt 2 $
So the modulus of $z = 1 + i$ is $\sqrt 2 $.
So $z = 1 + i$ can be represented as $z = \sqrt 2 {e^{i\dfrac{\pi }{4}}}$.
Now rotate the line OP in the anticlockwise direction by an angle of $\dfrac{\pi }{2}$ to get a new complex number represented by ${z_2}$.
Since we know that, if a complex number $z$ is rotated by an angle $\theta $ in the anticlockwise direction, then we get the answer to be $z{e^{i\theta }}$.
If $z = \sqrt 2 {e^{i\dfrac{\pi }{4}}}$ is rotated by an angle $\theta = \dfrac{\pi }{2}$ in the anticlockwise direction, then we get the answer to be ${z_2} = z{e^{i\theta }}$.
$
  {z_2} = \sqrt 2 {e^{i\dfrac{\pi }{4}}}{e^{i\dfrac{\pi }{2}}} \\
  {z_2} = \sqrt 2 {e^{i\left( {\dfrac{\pi }{4} + \dfrac{\pi }{2}} \right)}} \\
  {z_2} = \sqrt 2 {e^{i\dfrac{{3\pi }}{4}}} \\
 $
Since, ${e^{i\theta }}$can be represented as $\cos \theta + i\sin \theta $
$
  {z_2} = \sqrt 2 \left( {\cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}} \right) \\
  {z_2} = \sqrt 2 \left( { - \dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right) \\
  {z_2} = - 1 + i \\
 $
Since, the complement of $z = a + ib$ is $\overline z = a - ib$.
For $z = 1 + i$, the complement is $\overline z = 1 - i$.
So,
\[
  {z_2} = - (1 - i) \\
  {z_2} = - (\overline z ) \\
 \]

So, the correct answer is “Option c”.

Note:
The complex number of the form $z = a + ib$, the angle made by this complex number with the real axis is determined by $\theta = {\tan ^{ - 1}}\dfrac{b}{a}$ and the modulus of the given complex number is given by $|z| = r = \sqrt {{a^2} + {b^2}} $.These values are used to write the given complex number in the form of $r(\cos \theta + i\sin \theta )$.