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Question

(A) Let $z = 4 - 3i$.The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{4 - 3i}$.To simplify,multiply the numerator and de

Vedclass · Accepted Answer

$\frac{4}{25} + \frac{3}{25}i$

Vedclass · Answer

(A) Let $z = 4 - 3i$.The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{4 - 3i}$.To simplify,multiply the numerator and denominator by the conjugate $(4 + 3i)$:$z^{-1} = \frac{1(4 + 3i)}{(4 - 3i)(4 + 3i)}$.Using the identity $(a - b)(a + b) = a^2 - b^2$:$z^{-1} = \frac{4 + 3i}{4^2 - (3i)^2} = \frac{4 + 3i}{16 - 9i^2}$.Since $i^2 = -1$:$z^{-1} = \frac{4 + 3i}{16 + 9} = \frac{4 + 3i}{25}$.Thus,$z^{-1} = \frac{4}{25} + \frac{3}{25}i$.

Column-$I$	Column-$II$
$A. Z_1 Z_2$	$1. \text{imaginary axis}$
$B. Z_1 + Z_2 = 0$	$2. \text{Im}(-Z_2)$
$C. \text{Im}(Z_1)$	$3. \|Z_1\|^2$
$D. \text{Re}(Z_1)$	$4. \text{Re}(Z_2)$

Find the multiplicative inverse of the complex number $4-3i$.

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