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Find the multiplicative inverse of the complex numbers given.
\[\left( {4 - 3i} \right)\]

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Hint- If Z is a complex number, and then the multiplicative inverse of the complex number is given by
${z^{ - 1}} = \overline {\dfrac{z}{{{{\left| z \right|}^2}}}} $ . Where z is a complex number of the form \[a + ib\] and its conjugate is \[a - ib\] .

Complete step-by-step solution -
Let $z = 4 - 3i$
As we know that to find the conjugate of a number, we replace i by –i.
Then $\overline z = 4 + 3i$
Now, we have to find the magnitude of z
As we know that if \[z{\text{ }} = a + ib\] then ,
$\left| z \right| = \sqrt {{a^2} + {b^2}} $
$
  \therefore \left| z \right| = \sqrt {{4^2} + {{( - 3)}^2}} \\
   \Rightarrow \left| z \right| = \sqrt {16 + 9} \\
   \Rightarrow \left| z \right| = 5 \\
 $
Therefore, the multiplicative inverse of z is given by
${z^{ - 1}} = \overline {\dfrac{z}{{{{\left| z \right|}^2}}}} $
Substituting the value of $\overline z {\text{ and }}\left| z \right|$ in the above equation, we get
\[
   \Rightarrow {z^{ - 1}} = \dfrac{{4 + 3i}}{{{5^2}}} \\
   \Rightarrow {z^{ - 1}} = \dfrac{4}{{25}} + \dfrac{3}{{25}}i \\
 \]
Hence, the multiplicative inverse of z is \[{z^{ - 1}} = \dfrac{4}{{25}} + \dfrac{3}{{25}}i\] .

Note- The number in the form of \[a + ib\] is known as complex numbers where a is the real part and b is the imaginary part. The above question can also be solved by writing the \[a + ib\] in reciprocal form and multiply and divide it with the conjugate of \[a + ib\] . After simplifying we will get the multiplicative inverse of the given complex number. The same way we do with the real numbers.

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