Question

The multiplicative inverse of $\dfrac{{3 + 4i}}{{4 - 5i}}$ is
$\left( 1 \right)$ $\left( {\dfrac{{ - 8}}{{25}},\dfrac{{31}}{{25}}} \right)$
$\left( 2 \right)$ $\left( {\dfrac{{ - 8}}{{25}},\dfrac{{ - 31}}{{25}}} \right)$
$\left( 3 \right)$ $\left( {\dfrac{8}{{25}},\dfrac{{ - 31}}{{25}}} \right)$
$\left( 4 \right)$ $\left( {\dfrac{{ - 8}}{{25}},\dfrac{{31}}{5}} \right)$

Answer 1

We have to find the multiplicative inverse of the given expression of the complex number . We solve this question using the properties of complex numbers . We should have the knowledge of values of iota function for various powers . We should also have the knowledge of how to rationalise a complex number . First , we will rationalise the given complex numbers . Then we will simplify the given expression , further for the multiplicative inverse of the expression , we will take the reciprocal of the expression and again solve the expression by rationalising . And hence we would get the value of the expression in terms of the given points by taking the value of the real part and complex part separately .

Complete step by step answer:
Given :
The multiplicative inverse of $\dfrac{{3 + 4i}}{{4 - 5i}}$ .
Let us consider that the complex number is such that :
$z = \dfrac{{3 + 4i}}{{4 - 5i}}$
Now , according to the question we have to find the value of ${z^{ - 1}}$ .
On rationalising $z$ , we can write the expression as :
$z = \dfrac{{3 + 4i}}{{4 - 5i}} \times \dfrac{{4 + 5i}}{{4 + 5i}}$
Now , we also know that formula for difference of squares of two numbers is given as :
${a^2} - {b^2} = (a + b)(a - b)$
Using the formula and expanding the terms of the numerator , we can write the expression as :
$z = \dfrac{{12 + 16i + 15i + 20{i^2}}}{{{4^2} - {{\left( {5i} \right)}^2}}}$
$z = \dfrac{{12 + 31i + 20{i^2}}}{{16 - 25{i^2}}}$
Now , we also know that the values for different powers of iota is given as :
$i = \sqrt { - 1}$
${i^2} = - 1$
Putting the value of ${i^2}$ in the expression , we can write the expression as :
$z = \dfrac{{12 + 31i + 20\left( { - 1} \right)}}{{16 - 25\left( { - 1} \right)}}$
$z = \dfrac{{12 + 31i - 20}}{{16 + 25}}$
On solving , we get the expression as :
$z = \dfrac{{ - 8 + 31i}}{{41}}$
We also know that the inverse of a function can be written as :
${z^{ - 1}} = \dfrac{1}{z}$
Now we find the value of multiplicative inverse of $z$ as :
${z^{ - 1}} = \dfrac{{41}}{{ - 8 + 31i}}$
Again on rationalising ${z^{ - 1}}$ , we can write the expression as :
${z^{ - 1}} = \dfrac{{41}}{{ - 8 + 31i}} \times \dfrac{{ - 8 - 31i}}{{ - 8 - 31i}}$
Again using the formula of difference of squares of two numbers and expanding the numerator , we can write the expression as :
${z^{ - 1}} = \dfrac{{41 \times \left( { - 8 - 31i} \right)}}{{{{\left( { - 8} \right)}^2} - {{\left( {31i} \right)}^2}}}$
${z^{ - 1}} = \dfrac{{41 \times \left( { - 8 - 31i} \right)}}{{64 - 961{i^2}}}$
Now using the values for different powers of iota , we can write the expression as :
${z^{ - 1}} = \dfrac{{41 \times \left( { - 8 - 31i} \right)}}{{64 + 961}}$
${z^{ - 1}} = \dfrac{{41 \times \left( { - 8 - 31i} \right)}}{{1025}}$
Now on cancelling the terms , we can write the expression as :
${z^{ - 1}} = \dfrac{{ - 8 - 31i}}{{25}}$
The expression can also be written as :
${z^{ - 1}} = \dfrac{{ - 8}}{{25}} + \dfrac{{ - 31i}}{{25}}$
Now , we can also write the complex numbers in form of points as :
$\left( {x,y} \right)$ where $x$ is the value of the real part of the complex number and $y$ is the value of the complex part of the complex number .
Hence , the value of the multiplicative inverse is $\left( {\dfrac{{ - 8}}{{25}},\dfrac{{ - 31}}{{25}}} \right)$ .

So, the correct answer is “Option 2”.

Note:
We could also solve this question by taking the reciprocal of the given expression in the first step and then following the same steps as done above of rationalising and putting the values of different powers of iota we will get the required answer .