Reciprocal of a Complex Number
The reciprocal of a complex number \( z = a + bi \) is another complex number \( 1/z \), defined as: $$ \frac{1}{z} = \frac{1}{a+bi} $$ such that their product equals 1: $$ z \cdot \frac{1}{z} = 1 $$ Here, 1 is the multiplicative identity. More generally, the reciprocal of a complex number follows the rule: $$ z^{-n} = \frac{1}{z^n} $$
To determine the reciprocal of a complex number \( z_1 \), we start with \( \frac{1}{z_1} \) and apply a sequence of algebraic transformations.
One standard approach to computing \( \frac{1}{z_1} \) is to multiply both the numerator and denominator by the complex conjugate \( \bar{z_1} \):
$$ z_2 = \frac{1}{ z_1 } \cdot \frac{\bar{z_1} }{ \bar{z_1} } $$
The resulting value, \( z_2 \), is the reciprocal of \( z_1 \), satisfying the fundamental property:
$$ z_1 \cdot z_2 = 1 $$
Alternative Method
If \( z_1 \neq 0 \), its reciprocal can also be computed directly using the formula:
$$ z_2 = \frac{1}{z_1} = \frac{\overline{z_1}}{|z_1|^2} $$
Here, \( \overline{z_1} \) is the complex conjugate of \( z_1 \), and \( |z_1|^2 \) is the squared modulus, given by \( z_1 \cdot \overline{z_1} \).
Personally, I find this method faster than the previous one—provided you remember the formula.
For example, if \( z_1 = a + bi \), where \( a, b \in \mathbb{R} \), its reciprocal is: $$ z_2 = \frac{a - bi}{a^2 + b^2} $$
Worked Example
Let’s apply this to a specific complex number:
$$ z = 3+5i $$
The reciprocal of \( z \) is:
$$ \frac{1}{z} = \frac{1}{3+5i} $$
To verify, we compute the product \( z \cdot (1/z) \):
$$ z \cdot \frac{1}{z} = (3+5i) \cdot \frac{1}{3+5i} $$
$$ z \cdot \frac{1}{z} = \frac{3+5i}{3+5i} $$
This is a division of two complex numbers. The standard technique for simplifying such expressions is to multiply the numerator and denominator by the conjugate of the denominator:
$$ z \cdot \frac{1}{z} = \frac{3+5i}{3+5i} \cdot \frac{3-5i}{3-5i} $$
Expanding both the numerator and denominator:
$$ z \cdot \frac{1}{z} = \frac{(3+5i) \cdot (3-5i)}{(3+5i) \cdot (3-5i)} $$
$$ z \cdot \frac{1}{z} = \frac{9 - 15i + 15i - 25i^2}{9 - 15i + 15i - 25i^2} $$
$$ z \cdot \frac{1}{z} = \frac{9 - 25i^2}{9 - 25i^2} $$
Since we know that \( i^2 = -1 \):
$$ z \cdot \frac{1}{z} = \frac{9 - 25(-1)}{9 - 25(-1)} $$
$$ z \cdot \frac{1}{z} = \frac{9 + 25}{9 + 25} $$
At this stage, both the numerator and denominator are real numbers:
$$ z \cdot \frac{1}{z} = \frac{34}{34} $$
Since this is a division of two identical real numbers, the result is simply:
$$ z \cdot \frac{1}{z} = 1 $$
Note: We could also have used the general formula for dividing two complex numbers:
$$ \frac{a+bi}{c+di} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2} i $$
Substituting \( a = 3 \), \( b = 5 \), \( c = 3 \), and \( d = 5 \):
$$ \frac{3+5i}{3+5i} = \frac{3 \cdot 3+5 \cdot 5}{3^2+5^2} + \frac{5 \cdot 3-3 \cdot 5}{3^2+5^2} i $$
$$ = \frac{9+25}{9+25} + \frac{15-15}{3^2+5^2} i $$
$$ = \frac{34}{34} + \frac{0}{3^2+5^2} i $$
$$ = 1 + 0i $$
$$ = 1 $$
Although the result is the same, I find the direct conjugate method easier to remember. That’s why I prefer computing the reciprocal of a complex number by multiplying the numerator and denominator by the conjugate of the denominator.
Example 2
Let's find the reciprocal of the complex number \( z_1 = 2 + 3i \).
The reciprocal \( z_2 \) of a nonzero complex number \( z_1 \) is given by the formula:
$$ z_2 = \frac{1}{z_1} = \frac{1}{2+3i} $$
To rationalize the denominator, we multiply both the numerator and denominator by the complex conjugate of \( z_1 \). The conjugate of \( 2 + 3i \) is \( 2 - 3i \), so:
$$ z_2 = \frac{1}{2 + 3i} \cdot \frac{2 - 3i}{2 - 3i} $$
$$ z_2 = \frac{2 - 3i}{(2 + 3i)(2 - 3i)} $$
Since the product of a complex number and its conjugate equals the square of its modulus, we simplify the denominator:
$$ z_2 = \frac{2 - 3i}{2^2 + 3^2} $$
$$ z_2 = \frac{2 - 3i}{4 + 9} $$
$$ z_2 = \frac{2 - 3i}{13} $$
Expressing this in standard form:
$$ z_2 = \frac{2}{13} - \frac{3}{13}i $$
Thus, the reciprocal of \( 2 + 3i \) is:
$$ z_2 = \frac{2 - 3i}{13} $$
Verification: To confirm the result, we compute the product \( z_1 \cdot z_2 \).
$$ z_1 \cdot z_2 = (2+3i) \cdot \frac{2 - 3i}{13} $$
$$ z_1 \cdot z_2 = \frac{(2+3i) \cdot (2-3i)}{13} $$
Expanding the numerator:
$$ z_1 \cdot z_2 = \frac{4 - 6i + 6i - 9i^2}{13} $$
$$ z_1 \cdot z_2 = \frac{4 - 9i^2}{13} $$
Since \( i^2 = -1 \), we substitute:
$$ z_1 \cdot z_2 = \frac{4 - 9(-1)}{13} $$
$$ z_1 \cdot z_2 = \frac{4 + 9}{13} $$
$$ z_1 \cdot z_2 = \frac{13}{13} $$
$$ z_1 \cdot z_2 = 1 $$
Since the product equals 1, the multiplicative identity, this confirms that \( z_2 \) is indeed the reciprocal of \( z_1 \).
Example 3 (Alternative and Faster Method)
Now, let's compute the reciprocal of \( z_1 = 2 + 3i \) using a more direct approach.
We apply the formula:
$$ z_2 = \frac{1}{z_1} = \frac{\overline{z_1}}{|z_1|^2} $$
The conjugate of \( z_1 \) is \( \overline{z_1} = 2 - 3i \), so:
$$ z_2 = \frac{2 - 3i}{|z_1|^2} $$
The squared modulus of \( z_1 \) is:
$$ |z_1|^2 = (\sqrt{2^2 + (-3)^2})^2 = 4 + 9 = 13 $$
Substituting:
$$ z_2 = \frac{2 - 3i}{13} $$
With this approach, we arrive at the reciprocal of \( z_1 \) more efficiently than in the previous example.
Proof
Consider a complex number \( z \) and its reciprocal \( \frac{1}{z} \):
$$ z = a + bi $$
$$ \frac{1}{z} = \frac{1}{a+bi} $$
We verify that multiplying \( z \) by its reciprocal yields 1:
$$ z \cdot \frac{1}{z} = (a+bi) \cdot \frac{1}{a+bi} $$
$$ z \cdot \frac{1}{z} = \frac{a+bi}{a+bi} $$
To simplify this expression, we multiply both the numerator and denominator by the complex conjugate of \( a + bi \), which is \( a - bi \):
$$ z \cdot \frac{1}{z} = \frac{a+bi}{a+bi} \cdot \frac{a-bi}{a-bi} $$
$$ z \cdot \frac{1}{z} = \frac{(a+bi)(a-bi)}{(a+bi)(a-bi)} $$
Expanding both the numerator and denominator:
$$ z \cdot \frac{1}{z} = \frac{a^2 - abi + abi - b^2i^2}{a^2 - abi + abi - b^2i^2} $$
$$ z \cdot \frac{1}{z} = \frac{a^2 - b^2i^2}{a^2 - b^2i^2} $$
Since \( i^2 = -1 \), we substitute:
$$ z \cdot \frac{1}{z} = \frac{a^2 - b^2(-1)}{a^2 - b^2(-1)} $$
$$ z \cdot \frac{1}{z} = \frac{a^2 + b^2}{a^2 + b^2} $$
Since the numerator and denominator are identical real numbers, their quotient is 1:
$$ z \cdot \frac{1}{z} = 1 $$
Properties of Complex Reciprocals
Some useful properties of the reciprocals of complex numbers:
- The reciprocal of \( z = a + bi \) is given by: $$ \frac{1}{z} = \frac{a - bi}{a^2 + b^2} $$
Proof: Starting from the definition of the reciprocal:
$$ \frac{1}{z} = \frac{1}{a+bi} $$
Multiplying the numerator and denominator by the conjugate \( \overline{z} = a - bi \):
$$ \frac{1}{z} = \frac{1}{a+bi} \cdot \frac{a - bi}{a - bi} $$
Expanding the denominator:
$$ \frac{1}{z} = \frac{a - bi}{(a+bi)(a-bi)} $$
$$ \frac{1}{z} = \frac{a - bi}{a^2 - abi + abi - b^2i^2} $$
$$ \frac{1}{z} = \frac{a - bi}{a^2 - b^2i^2} $$
Since \( i^2 = -1 \), we simplify:
$$ \frac{1}{z} = \frac{a - bi}{a^2 - b^2(-1)} $$
$$ \frac{1}{z} = \frac{a - bi}{a^2 + b^2} $$
Verification: To verify, we multiply the reciprocal by \( z = a + bi \):
$$ z \cdot \frac{1}{z} = (a+bi) \cdot \frac{a - bi}{a^2 + b^2} $$
$$ z \cdot \frac{1}{z} = \frac{(a+bi)(a - bi)}{a^2 + b^2} $$
Expanding:
$$ z \cdot \frac{1}{z} = \frac{a^2 - abi + abi - b^2i^2}{a^2 + b^2} $$
$$ z \cdot \frac{1}{z} = \frac{a^2 - b^2i^2}{a^2 + b^2} $$
Again, substituting \( i^2 = -1 \):
$$ z \cdot \frac{1}{z} = \frac{a^2 - b^2(-1)}{a^2 + b^2} $$
$$ z \cdot \frac{1}{z} = \frac{a^2 + b^2}{a^2 + b^2} $$
Since this simplifies to 1, the formula is confirmed.
Reciprocal of a Complex Number in Trigonometric Form
The reciprocal of a complex number expressed in trigonometric form, \( z = r(\cos \alpha + i \sin \alpha) \), is given by:
$$ \frac{1}{z} = \frac{1}{r} \cdot (\cos \alpha - i \sin \alpha) $$
More generally, for negative exponents:
$$ z^{-n} = \frac{1}{r^n} \cdot (\cos n\alpha - i \sin n\alpha) $$
Computing the reciprocal is often more straightforward when the complex number is in trigonometric form.
A Practical Example
Consider the complex number:
$$ z = 3+5i $$
We express it in trigonometric form:
$$ z = \sqrt{3^2+5^2} \cdot \left[ \cos \left( \arctan \frac{5}{3} \right) + i \sin \left( \arctan \frac{5}{3} \right) \right] $$
$$ z = \sqrt{34} \cdot [ \cos ( 59.04^\circ ) + i \sin ( 59.04^\circ ) ] $$
Applying the reciprocal formula:
$$ \frac{1}{z} = \frac{1}{\sqrt{34}} \cdot [ \cos(59.04^\circ) - i \sin(59.04^\circ)] $$
Converting back to standard form:
$$ x = \frac{1}{\sqrt{34}} \cdot \cos (59.04^\circ) = 0.09 $$
$$ y = \frac{1}{\sqrt{34}} \cdot [-\sin (59.04^\circ)] = -0.15 $$
Thus, the reciprocal in rectangular form is:
$$ \frac{1}{z} = 0.09 - 0.15i $$
