Find the multiplicative inverse of 2-3i. | vedclass

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Question

(A) Let $z = 2-3i$.The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{2-3i}$.To simplify,multiply the numerator and denomi

Vedclass · Accepted Answer

$\frac{2}{13}+\frac{3}{13}i$

Vedclass · Answer

(A) Let $z = 2-3i$.The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{2-3i}$.To simplify,multiply the numerator and denominator by the conjugate of $z$,which is $2+3i$:$z^{-1} = \frac{1(2+3i)}{(2-3i)(2+3i)}$Using the identity $(a-bi)(a+bi) = a^2+b^2$:$z^{-1} = \frac{2+3i}{2^2 + (-3)^2} = \frac{2+3i}{4+9} = \frac{2+3i}{13}$Thus,$z^{-1} = \frac{2}{13} + \frac{3}{13}i$.

Find the multiplicative inverse of $2-3i$.

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