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Question

Mathematics Question on complex numbers

The multiplicative inverse of 3+4i4−5i \frac{3 + 4i}{4 - 5 i} is

A

(−825,3125)\left(\frac{-8}{25} , \frac{31}{25} \right)

B

(−825,−3125)\left(\frac{-8}{25} , \frac{-31}{25} \right)

C

(825,−3125)\left(\frac{8}{25} , \frac{-31}{25} \right)

D

(825,3125)\left(\frac{8}{25} , \frac{31}{25} \right)

Answer

(−825,−3125)\left(\frac{-8}{25} , \frac{-31}{25} \right)

Explanation

Solution

Let z=3+4i4−5iz=\frac{3+4i}{4-5i}
we have to calculate z−1 i.e.,1zz^{-1} \, i.e., \frac{1}{z}
∴   z−1=1z=4−5i3+4i×3−4i3−4i\therefore \:\:\: z^{-1} =\frac{1}{z} = \frac{4-5i}{3+4i}\times\frac{3-4i}{3-4i}
=12−15i−16i+20i29−16i2=12−31i−209−16(−1)= \frac{12-15i -16i+20i^{2}}{9-16i^{2}} =\frac{12-31i-20}{9-16\left(-1\right)}
=−8−31i9+16=−825−3125i= \frac{-8-31i}{9+16}=\frac{-8}{25} -\frac{31}{25}i
∴    z−1=(−825,−3125)\therefore\:\:\:\: z^{-1}=\left( \frac{-8}{25} , \frac{-31}{25}\right)