Question

In the argand diagram, if \[O\], \[P\], and \[Q\] represents respectively the origin, the complex numbers \[z,\text{ }z+iz\], then the angle $\angle OPQ$ is
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{2}$
D. $\dfrac{2\pi }{3}$

Answer 1

Hint: In this question, we have to find the angle at \[P\] i.e., $\angle OPQ$. For this, we can use the slopes of the lines \[\overrightarrow{OP}\] and \[\overrightarrow{PQ}\]. This is the fundamental concept of angles.

Formula Used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: It is given that, in an argand plane, \[O\], \[P\], and \[Q\] represents respectively the origin, the complex numbers \[z,\text{ }z+iz\].
So, we can write them in the coordinate form as
\[\begin{align}
  & O=0+0i=(0,0) \\
 & P=z=x+iy=(x,y) \\
\end{align}\]
\[\begin{align}
  & Q=z+iz \\
 & \text{ }=(x+iy)+i(x+iy) \\
 & \text{ }=x+iy+ix-y \\
 & \text{ }=(x-y)+i(x+y) \\
 & \text{ }=\left( (x-y),(x+y) \right) \\
\end{align}\]
Then, the slope of the line \[\overrightarrow{OP}\] is
$\begin{align}
  & {{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\dfrac{y-0}{x-0} \\
 & \text{ }=\dfrac{y}{x} \\
\end{align}$
So, the slope of the line \[\overrightarrow{PQ}\] is
$\begin{align}
  & {{m}_{2}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\dfrac{(x+y)-y}{(x-y)-x} \\
 & \text{ }=\dfrac{x}{-y} \\
\end{align}$
On multiplying both the slopes we get,
${{m}_{1}}\times {{m}_{2}}=\dfrac{y}{x}\times \dfrac{x}{-y}=-1$
Therefore, the required angle $\angle OPQ$ is $\dfrac{\pi }{2}$.

Option ‘C’ is correct

Note: Here we need to remember that, if the product of slopes of any two lines is equal to $-1$, then those lines are said to be perpendicular to each other. By using this, we can able to calculate the angle. If it is not equal to $-1$, then we can use the formula for finding the angle between two lines.