
If the points ${{P}_{1}}$ and ${{P}_{2}}$ represent two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$, then the point ${{P}_{3}}$ represents

A. ${{z}_{1}}+{{z}_{2}}$
B. ${{z}_{1}}-{{z}_{2}}$
C. ${{z}_{1}}\times {{z}_{2}}$
D. ${{z}_{1}}\div {{z}_{2}}$
Answer
228k+ views
Hint: In this question, we have to find the complex number at point ${{P}_{3}}$. For this, the property of parallelogram i.e., the midpoints of the diagonals of a parallelogram are equal is used.
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: Given that,
The points ${{P}_{1}}$ and ${{P}_{2}}$ are represented by the complex numbers ${{z}_{1}}$ and ${{z}_{2}}$.
These are the vertices of the given parallelogram $O{{P}_{1}}{{P}_{2}}{{P}_{3}}$
So, the midpoints of its diagonals are equal. I.e.,
${{P}_{1}}{{P}_{2}}=O{{P}_{3}}$
The midpoint of ${{P}_{1}}{{P}_{2}}$ is
$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\text{ }...(1)$
The midpoint of $O{{P}_{3}}$ is
$\left( \dfrac{0+x}{2},\dfrac{0+y}{2} \right)\text{ }...(2)$
On equating (1) and (2), we get
\[{{P}_{3}}(x,y)=({{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}})\]
Thus, this point is represented in the complex form as
$\begin{align}
& ({{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}})=({{x}_{1}}+{{x}_{2}})+i({{y}_{1}}+{{y}_{2}}) \\
& \Rightarrow ({{x}_{1}}+i{{y}_{1}})+({{x}_{2}}+i{{y}_{2}}) \\
& \Rightarrow {{z}_{1}}+{{z}_{2}} \\
\end{align}$
Thus, the point ${{P}_{3}}$ is represented by ${{z}_{1}}+{{z}_{2}}$.
Option ‘A’ is correct
Note: Here we need to remember that, the required point is one of the vertices of the given parallelogram. So, the property of the parallelogram is applied to find the required vertex.
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: Given that,
The points ${{P}_{1}}$ and ${{P}_{2}}$ are represented by the complex numbers ${{z}_{1}}$ and ${{z}_{2}}$.
These are the vertices of the given parallelogram $O{{P}_{1}}{{P}_{2}}{{P}_{3}}$
So, the midpoints of its diagonals are equal. I.e.,
${{P}_{1}}{{P}_{2}}=O{{P}_{3}}$
The midpoint of ${{P}_{1}}{{P}_{2}}$ is
$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\text{ }...(1)$
The midpoint of $O{{P}_{3}}$ is
$\left( \dfrac{0+x}{2},\dfrac{0+y}{2} \right)\text{ }...(2)$
On equating (1) and (2), we get
\[{{P}_{3}}(x,y)=({{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}})\]
Thus, this point is represented in the complex form as
$\begin{align}
& ({{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}})=({{x}_{1}}+{{x}_{2}})+i({{y}_{1}}+{{y}_{2}}) \\
& \Rightarrow ({{x}_{1}}+i{{y}_{1}})+({{x}_{2}}+i{{y}_{2}}) \\
& \Rightarrow {{z}_{1}}+{{z}_{2}} \\
\end{align}$
Thus, the point ${{P}_{3}}$ is represented by ${{z}_{1}}+{{z}_{2}}$.
Option ‘A’ is correct
Note: Here we need to remember that, the required point is one of the vertices of the given parallelogram. So, the property of the parallelogram is applied to find the required vertex.
Recently Updated Pages
What is the solution of the differential equation edfracdydx class 11 maths JEE_Advanced

In a GP of 3rmn terms S1 denotes the sum of first rmn class 11 maths JEE_Advanced

In an isosceles triangle ABC the coordinates of the class 11 maths JEE_Advanced

If a1+left sqrt31 right+left sqrt31 right2+left sqrt31 class 11 maths JEE_Advanced

What is the value of the integral intlimits 11 sin class 11 maths JEE_Advanced

int 44 left x + 2 right dx A 50 B 24 C 20 D None of class 11 maths JEE_Advanced

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2026 Revision Notes for Practical Organic Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

IIT Fees Structure 2025

Other Pages
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

