How to find the reciprocal of a complex number?
Let z = x + iy be a non-zero complex number. Then
\(\frac{1}{z}\)
= \(\frac{1}{x + iy}\)
= \(\frac{1}{x + iy}\) × \(\frac{x - iy}{x - iy}\), [Multiplying numerator and denominator by conjugate of denominator i.e., Multiply both numerator and denominator by conjugate of x + iy]
= \(\frac{x - iy}{x^{2} - i^{2}y^{2}}\)
= \(\frac{x - iy}{x^{2} + y^{2}}\)
= \(\frac{x}{x^{2} + y^{2}}\) + \(\frac{i(-y)}{x^{2} + y^{2}}\)
Clearly, \(\frac{1}{z}\) is equal to the multiplicative inverse of z. Also,
\(\frac{1}{z}\) = \(\frac{x - iy}{x^{2} + y^{2}}\) = \(\frac{\overline{z}}{|z|^{2}}\)
Therefore, the multiplicative inverse of a non-zero complex z is equal to its reciprocal and is represent as
\(\frac{Re(z)}{|z|^{2}}\) + i\(\frac{(-Im(z))}{|z|^{2}}\)= \(\frac{\overline{z}}{|z|^{2}}\)
Solved examples on reciprocal of a complex number:
1. If a complex number z = 2 + 3i, then find the reciprocal of z? Give your answer in a + ib form.
Solution:
Given z = 2 + 3i
Then, \(\overline{z}\) = 2 - 3i
And |z| = \(\sqrt{x^{2} + y^{2}}\)
= \(\sqrt{2^{2} + (-3)^{2}}\)
= \(\sqrt{4 + 9}\)
= \(\sqrt{13}\)
Now, |z|\(^{2}\) = 13
Therefore, \(\frac{1}{z}\) = \(\frac{\overline{z}}{|z|^{2}}\) = \(\frac{2 - 3i}{13}\) = \(\frac{2}{13}\) + (-\(\frac{3}{13}\))i, which is the required a + ib form.
2. Find the reciprocal of the complex number z = -1 + 2i. Give your answer in a + ib form.
Solution:
Given z = -1 + 2i
Then, \(\overline{z}\) = -1 - 2i
And |z| = \(\sqrt{x^{2} + y^{2}}\)
= \(\sqrt{(-1)^{2} + 2^{2}}\)
= \(\sqrt{1 + 4}\)
= \(\sqrt{5}\)
Now, |z|\(^{2}\)= 5
Therefore, \(\frac{1}{z}\) = \(\frac{\overline{z}}{|z|^{2}}\) = \(\frac{-1 - 2i}{5}\) = (-\(\frac{1}{5}\)) + (-\(\frac{2}{5}\))i, which is the required a + ib form.
3. Find the reciprocal of the complex number z = i. Give your answer in a + ib form.
Solution:
Given z = i
Then, \(\overline{z}\) = -i
And |z| = \(\sqrt{x^{2} + y^{2}}\)
= \(\sqrt{0^{2} + 1^{2}}\)
= \(\sqrt{0 + 1}\)
= \(\sqrt{1}\)
= 1
Now, |z|\(^{2}\)= 1
Therefore, \(\frac{1}{z}\) = \(\frac{\overline{z}}{|z|^{2}}\) = \(\frac{-i}{1}\) = -i = 0 + (-i), which is the required a + ib form.
Note: The reciprocal of i is its own conjugate - i.
11 and 12 Grade Math
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