Question

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

Answer 1

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.