Reciprocal of a Complex Number

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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