Question:
The multiplicative inverse of $ \frac{3 + 4i}{4 - 5 i}$ is
The multiplicative inverse of $ \frac{3 + 4i}{4 - 5 i}$ is
Updated On: May 11, 2024
- $\left(\frac{-8}{25} , \frac{31}{25} \right)$
- $\left(\frac{-8}{25} , \frac{-31}{25} \right)$
- $\left(\frac{8}{25} , \frac{-31}{25} \right)$
- $\left(\frac{8}{25} , \frac{31}{25} \right)$
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The Correct Option is B
Solution and Explanation
Let $z=\frac{3+4i}{4-5i}$
we have to calculate $z^{-1} \, i.e., \frac{1}{z}$
$ \therefore \:\:\: z^{-1} =\frac{1}{z} = \frac{4-5i}{3+4i}\times\frac{3-4i}{3-4i}$
$ = \frac{12-15i -16i+20i^{2}}{9-16i^{2}} =\frac{12-31i-20}{9-16\left(-1\right)} $
$= \frac{-8-31i}{9+16}=\frac{-8}{25} -\frac{31}{25}i$
$\therefore\:\:\:\: z^{-1}=\left( \frac{-8}{25} , \frac{-31}{25}\right)$
we have to calculate $z^{-1} \, i.e., \frac{1}{z}$
$ \therefore \:\:\: z^{-1} =\frac{1}{z} = \frac{4-5i}{3+4i}\times\frac{3-4i}{3-4i}$
$ = \frac{12-15i -16i+20i^{2}}{9-16i^{2}} =\frac{12-31i-20}{9-16\left(-1\right)} $
$= \frac{-8-31i}{9+16}=\frac{-8}{25} -\frac{31}{25}i$
$\therefore\:\:\:\: z^{-1}=\left( \frac{-8}{25} , \frac{-31}{25}\right)$
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.