Answer

Verified

447.9k+ views

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.

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. \[\therefore {{y}_{1}}+{{y}_{2}}=0\]

__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. \[\therefore {{y}_{1}}+{{y}_{2}}=0\]

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE