Equality of Complex Numbers

We will discuss about the equality of complex numbers.

Two complex numbers z\(_{1}\) = a + ib and z\(_{2}\) = x + iy are equal if and only if a = x and b = y i.e., Re (z\(_{1}\)) = Re (z\(_{2}\)) and Im (z\(_{1}\)) = Im (z\(_{2}\)).

Thus, z\(_{1}\) = z\(_{2}\) ⇔ Re (z\(_{1}\)) = Re (z\(_{2}\)) and Im (z\(_{1}\)) = Im (z\(_{2}\)).

For example, if the complex numbers z\(_{1}\) = x + iy and z\(_{2}\) = -5 + 7i are equal, then x = -5 and y = 7.


Solved examples on equality of two complex numbers:

1. If z\(_{1}\) = 5 + 2yi and z\(_{2}\) = -x + 6i are equal, find the value of x and y.

Solution:

The given two complex numbers are z\(_{1}\) = 5 + 2yi and z\(_{2}\) = -x + 6i.

We know that, two complex numbers z\(_{1}\) = a + ib and z\(_{2}\) = x + iy are equal if a = x and b = y.

z\(_{1}\) = z\(_{2}\)

⇒ 5 + 2yi = -x + 6i

⇒ 5 = -x and 2y = 6

⇒ x = -5 and y = 3

Therefore, the value of x = -5 and the value of y = 3.

 

2. If a, b are real numbers and 7a + i(3a - b) = 14 - 6i, then find the values of a and b.

Solution:

Given, 7a + i(3a - b) = 14 - 6i

⇒ 7a + i(3a - b) = 14 + i(-6)

Now equating real and imaginary parts on both sides, we have

7a = 14 and 3a - b = -6

⇒ a = 2 and 3 2 – b = -6

⇒ a = 2 and 6 – b = -6

⇒ a = 2 and – b = -12

⇒ a = 2 and b = 12

Therefore, the value of a = 2 and the value of b = 12.

 

3. For what real values of m and n are the complex numbers m\(^{2}\) – 7m + 9ni and n\(^{2}\)i + 20i -12 are equal.

Solution:

Given complex numbers are m\(^{2}\) - 7m + 9ni and n\(^{2}\)i + 20i -12

According to the problem,

m\(^{2}\) - 7m + 9ni = n\(^{2}\)i + 20i -12

⇒ (m\(^{2}\) - 7m) + i(9n) = (-12) + i(n\(^{2}\) + 20)

Now equating real and imaginary parts on both sides, we have

m\(^{2}\) - 7m = - 12 and 9n = n\(^{2}\) + 20

⇒ m\(^{2}\) - 7m + 12 = 0 and n\(^{2}\) - 9n + 20 = 0

⇒ (m - 4)(m - 3) = 0 and (n - 5)(n - 4) = 0

⇒ m = 4, 3 and n = 5, 4

Hence, the required values of m and n are follows:

m = 4, n = 5; m = 4, n = 4; m = 3, n = 5; m = 3, n = 4.




11 and 12 Grade Math 

From Equality of Complex Numbers to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Multiplication by Ten, Hundred and Thousand |Multiply by 10, 100 &1000

    Jan 17, 25 12:34 PM

    Multiply by 10
    To multiply a number by 10, 100, or 1000 we need to count the number of zeroes in the multiplier and write the same number of zeroes to the right of the multiplicand. Rules for the multiplication by 1…

    Read More

  2. Multiplying 2-Digit Numbers by 2-Digit Numbers |Multiplying by 2-Digit

    Jan 17, 25 01:46 AM

    Multiplying 2-Digit Numbers by 2-Digit Numbers
    We will learn how to multiply 2-digit numbers by 2-digit numbers.

    Read More

  3. Multiplying 3-Digit Numbers by 2-Digit Numbers | 3-Digit by 2-Digit

    Jan 17, 25 01:17 AM

    Multiplying 3-Digit Numbers by 2-Digit Numbers
    "We will learn how to multiply 3-digit numbers by 2-digit numbers.

    Read More

  4. 4-Digits by 1-Digit Multiplication |Multiply 4-Digit by 1-Digit Number

    Jan 17, 25 12:01 AM

    4-Digit by 1-Digit Multiply
    Here we will learn 4-digits by 1-digit multiplication. We know how to multiply three digit number by one digit number. In the same way we can multiply 4-digit numbers by 1-digit numbers without regrou…

    Read More

  5. Multiplying 3-Digit Number by 1-Digit Number | Three-Digit Multiplicat

    Jan 15, 25 01:54 PM

    Multiplying 3-Digit Number by 1-Digit Number
    Here we will learn multiplying 3-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. 1. Multiply 201 by 3 Step I: Arrange the numb…

    Read More