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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 \[\left| z \right| = \sqrt {{4^2} + {{\left( { - 3} \right)}^2}} \]

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 \[\left| z \right| = \sqrt {{4^2} + {{\left( { - 3} \right)}^2}} \]