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
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