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# Under what condition is the sum of two complex numbers ${{x}_{1}}+i{{y}_{1}}$ and ${{x}_{2}}+i{{y}_{2}}$ is a purely imaginary number?A. ${{x}_{1}}+{{x}_{2}}=0$B. ${{y}_{1}}+{{y}_{2}}=0$C. ${{x}_{1}}+{{x}_{2}}={{y}_{1}}+{{y}_{2}}$D. ${{x}_{1}}={{y}_{1}},{{x}_{2}}={{y}_{2}}$ Verified
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Hint: Add both the complex numbers formed and separate the real part and the imaginary part. To get a purely imaginary part, the real part should be zero.

Complete step-by-step solution -
A complex number is a number that can be expressed in the form of $a+ib$, where $a$ and $b$ are real numbers and $i$ is the solution of the equation ${{x}^{2}}=-1$. As no real number satisfies this equation, so $i$ is called an imaginary number.
For a complex number $a+ib$ ,$a$ is called the real part and $b$ is called the imaginary part.
A complex number system can be defined as the algebraic extension is the ordinary real number by an imaginary number. A complex number whose real part is zero can be called to be purely imaginary, the points for these numbers lie on the vertical axis of the complex plane.
Similarly, a complex number whose imaginary part is zero can be viewed as purely real, the points lie on the horizontal axis of the complex plane.
Given to us are the two complex numbers $\left( {{x}_{1}}+i{{y}_{1}} \right)$and$\left( {{x}_{2}}+i{{y}_{2}} \right)$.
In this, ${{x}_{1}}$ and ${{x}_{2}}$ are the real parts of the complex number. ${{y}_{1}}$ and ${{y}_{2}}$ are the imaginary parts of the complex number.
Thus if we are adding both complex numbers, we get,
$\left( {{x}_{1}}+i{{y}_{1}} \right)+\left( {{x}_{2}}+i{{y}_{2}} \right) = \left( {{x}_{1}}+{{x}_{2}} \right)+i\left( {{y}_{1}}+{{y}_{2}} \right)$.
In this expression formed $\left( {{x}_{1}}+{{x}_{2}} \right)$ is the real part and $i\left( {{y}_{1}}+{{y}_{2}} \right)$ is the imaginary part.
So if the sum has to be purely imaginary then the real part should be zero.
$\therefore {{x}_{1}}+{{x}_{2}}=0$, then the sum becomes purely imaginary.
Hence, option A is the correct answer.

Note:- If the problem was to find the purely real then $\left( {{y}_{1}}+{{y}_{2}} \right)$ should become zero as $\left( {{y}_{1}}+{{y}_{2}} \right)$ are the real numbers. So to get purely real numbers, $\left( {{y}_{1}}+{{y}_{2}} \right)$ should be zero. $\therefore {{y}_{1}}+{{y}_{2}}=0$

Last updated date: 28th Sep 2023
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