Solveeit Logo

Question

Question: How do find the product of \[7 - 2i\] and its conjugate....

How do find the product of 72i7 - 2i and its conjugate.

Explanation

Solution

This problem comes under complex numbers. Complex numbers means algebraic expression in term imaginary numbers say ii which can be expressed in the form a+iba + ib, where a and b are real numbers ii is imaginary number, i2=1{i^2} = - 1. In this problem we need to find the product of the conjugate of the given complex number. Then we find the solution with basic mathematical calculation.

Formula used: (x+iy)(xiy)=x2i2y2+ixyixy(x + iy)(x - iy) = {x^2} - {i^2}{y^2} + ixy - ixy
(x+iy)(xiy)=x2+y2(x + iy)(x - iy) = {x^2} + {y^2}

Complete step-by-step solution:
Let us consider a given complex number 72i7 - 2i
Now, the conjugate 72i7 - 2i is
In this the sign of the imaginary part of a given complex number is negative, so the conjugate will be the imaginary part will become negative and the real part remains the same.
7+2i\Rightarrow 7 + 2i
Now the product of given complex number with conjugate is
(72i)(7+2i)\Rightarrow (7 - 2i)(7 + 2i)
Now used formula mentioned in formula used, we get
7222i2\Rightarrow {7^2} - {2^2}{i^2}
We know that i2=1{i^2} = - 1 by simplifying it in the above equation and finding square of given number, we get
49+4\Rightarrow 49 + 4
On adding the term and we get,
53\Rightarrow 53

Therefore, the product of given complex number and its conjugate is 5353

Note: Complex numbers appear when an imaginary part of the equation comes that is negative integers in root value. We need to know about real and imaginary parts of complex numbers.
Conjugate in complex numbers means that the change of signs in the imaginary part of the complex. If a positive sign in an imaginary part is a complex number then the conjugate of complex numbers in the imaginary part is negative and vice versa.
The complex number expressed as a+iba + ib. Then to find the complex conjugate and with complex multiplication and properties we solve this. We need to be familiar with that so that it will be easy to solve.