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# Find the multiplicative inverse of the complex numbers $z = 4 - 3i$.

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Hint- The multiplicative inverse of complex number $z = 4 - 3i$ is given by ${z^{ - 1}} = \dfrac{{{\text{conjugate of z}}}}{{{{\left| z \right|}^{^2}}}}$.

Multiplicative Inverse of a non-zero complex number $z$ is an element denoted by ${z^{ - 1}}$ such that $z{z^{ - 1}} = 1$.
Method 1.
Now according to the question Let $z = 4 - 3i$and we know that multiplicative inverse is given as
${z^{ - 1}} = \dfrac{{{\text{conjugate of z}}}}{{{{\left| z \right|}^{^2}}}}$
And conjugate of $z = 4 - 3i$ is nothing but changing the sign of $i$ term i.e. the conjugate of $z =$$4 + 3i$
Where $\left| z \right|$ indicates the magnitude of complex number $z = 4 - 3i$ given as $\left| z \right| = \sqrt {{4^2} + {{\left( { - 3} \right)}^2}}$
So,
$\Rightarrow {\left| z \right|^2} = {4^2} + {\left( { - 3} \right)^2} \\ \Rightarrow {\left| z \right|^2} = 16 + 9 \\ \Rightarrow {\left| z \right|^2} = 25 \\$
Therefore, the multiplicative inverse of $4 - 3i$ is
${z^{ - 1}} = \dfrac{{{\text{conjugate of z}}}}{{{{\left| z \right|}^{^2}}}} = \dfrac{{4 + 3i}}{{25}} = \dfrac{4}{{25}} + \dfrac{3}{{25}}i$
Method 2.
Let the multiplicative inverse of $z = 4 - 3i$ is $x$
Then $x = \dfrac{1}{{4 - 3i}}$
Now Rationalize the denominator as
$x = \dfrac{1}{{4 - 3i}} \times \dfrac{{4 + 3i}}{{4 + 3i}}$
Using $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, then
$x = \dfrac{{4 + 3i}}{{{4^2} - {{\left( {3i} \right)}^2}}} = \dfrac{{4 + 3i}}{{16 + 9}}$
$x = \dfrac{{4 + 3i}}{{25}}$
the multiplicative inverse of $4 - 3i$ is $\dfrac{{4 + 3i}}{{25}}$.

Note- A Complex number in rectangular form is represented by $z = a + ib$. where $a$ is real part and $b$ is the imaginary part in an argand plane. conjugate of $z = 4 - 3i$ is nothing but changing the sign of $i$ term. $\left| z \right|$ indicates the magnitude of complex number $z = 4 - 3i$ given as $\left| z \right| = \sqrt {{4^2} + {{\left( { - 3} \right)}^2}}$