Certainly! To show that [tex]\(\frac{1}{z} = \frac{\bar{z}}{|z|^2}\)[/tex] for a nonzero complex number [tex]\( z = a + bi \)[/tex], we will proceed with the following steps:
1. Step 1: Definition of Conjugate
The conjugate of a complex number [tex]\( z = a + bi \)[/tex] is given by:
[tex]\[
\bar{z} = a - bi
\][/tex]
2. Step 2: Definition of Modulus and Modulus Squared
The modulus of [tex]\( z = a + bi \)[/tex] is:
[tex]\[
|z| = \sqrt{a^2 + b^2}
\][/tex]
Therefore, the modulus squared is:
[tex]\[
|z|^2 = (a + bi)(a - bi) = a^2 + b^2
\][/tex]
3. Step 3: Finding the Multiplicative Inverse
To find [tex]\(\frac{1}{z}\)[/tex], we multiply both numerator and denominator by the conjugate of [tex]\( z \)[/tex]:
[tex]\[
\frac{1}{z} = \frac{1}{a + bi} \cdot \frac{a - bi}{a - bi} = \frac{a - bi}{(a + bi)(a - bi)}
\][/tex]
Simplifying the denominator using the result from Step 2:
[tex]\[
\frac{1}{z} = \frac{a - bi}{a^2 + b^2}
\][/tex]
4. Step 4: Expressing the Result
Notice that [tex]\( \bar{z} = a - bi \)[/tex] and [tex]\( |z|^2 = a^2 + b^2 \)[/tex]. Then:
[tex]\[
\frac{1}{z} = \frac{a - bi}{a^2 + b^2} = \frac{\bar{z}}{|z|^2}
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{1}{z} = \frac{\bar{z}}{|z|^2}
\][/tex]
This completes the proof.
