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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,$z^{-1} = \frac{4}{25} + \frac{3}{25}i$.

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

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