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The multiplicative inverse of $\mathrm{z}$ is
Options:
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Verified Answer
The correct answer is:
$\frac{\bar{z}}{|z|^2}$
Let multiplicative inverse of $\mathrm{z}$, be 'A'
hence $\mathrm{z} \cdot \mathrm{A}=1$
$\Rightarrow \quad A=\frac{1}{z}=\frac{1 \cdot \bar{z}}{z \cdot \bar{z}} \quad$ (where $\bar{z}$ is conjugate of $z$ )
$\Rightarrow A=\frac{\bar{z}}{z \bar{z}}=\frac{\bar{z}}{|z|^2} \quad\left\{\because z \bar{z}=|z|^2\right\}$
Hence option (c) is correct.
hence $\mathrm{z} \cdot \mathrm{A}=1$
$\Rightarrow \quad A=\frac{1}{z}=\frac{1 \cdot \bar{z}}{z \cdot \bar{z}} \quad$ (where $\bar{z}$ is conjugate of $z$ )
$\Rightarrow A=\frac{\bar{z}}{z \bar{z}}=\frac{\bar{z}}{|z|^2} \quad\left\{\because z \bar{z}=|z|^2\right\}$
Hence option (c) is correct.
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