Question

Find the multiplicative inverse of the complex number $4-3i$.

Answer 1

Hint: As we know that the multiplicative inverse is reciprocal of a number. Suppose we have a number $\left( \dfrac{a}{b} \right)$ then its multiplicative inverse is $\left( \dfrac{b}{a} \right)$, so that after multiplying it to the number we get the multiplicative identity, i.e. 1.

Complete step-by-step answer:
We have been given the complex number which is $\left( 4-3i \right)$ and we have to find its multiplicative inverse.
We know that the multiplicative inverse is reciprocal of the given number. After that we will rationalize the reciprocal by multiplying both the numerator and denominator by its complex conjugate, which is another complex number with opposite sign between the real and imaginary part of the original number.
So, the multiplicative inverse of $\left( 4-3i \right)$ = $\dfrac{1}{4-3i}$.
Now the conjugate of $\left( 4-3i \right)$ is $\left( 4+3i \right)$. So, we can multiply both the numerator and denominator with the conjugate and we will get,
$\dfrac{1}{4-3i}=\dfrac{1}{4-3i}\times \dfrac{4+3i}{4+3i}=\dfrac{4+3i}{{{4}^{2}}-9{{i}^{2}}}$
And we know that the value of ${{i}^{2}}=-1$.
On substituting the value of ${{i}^{2}}=-1$ in the expression and then simplifying it further, we get,
$\dfrac{1}{4-3i}=\dfrac{4+3i}{16-9(-1)}=\dfrac{4+3i}{16+9}=\dfrac{1}{25}\left[ 4+3i \right]$
Hence the conjugate of the given complex number $\left( 4-3i \right)$ is equal to $\dfrac{1}{25}\left[ 4-3i \right]$.

Note: Don’t change the sign between the real and imaginary part while finding the imaginative inverse. Only find the reciprocal of the number. Also students must not stop by just finding the multiplicative inverse. They have to rationalise the denominator and then use formulas to express it in simplest terms possible. They should express it as a sum of real and complex parts, a+ib.