Question

solve the question with explanation..​

Answer 1

AnswEr :

Given Expression,

$\sf \: z = \dfrac{3 + 4 \imath}{4 - 5 \imath}$

We have to find the Multiplicative Inverse of z

The product of a number and it's multiplicative inverse would be :

$\sf \: z \times \dfrac{1}{z} = 1$

Let the Multiplicative Inverse be x

Thus,

$\longrightarrow \: \sf \: x = \dfrac{4 - 5 \imath}{3 + 4 \imath}$

Multiplying the numerator and denominator with 3 - 4i ( Conjugate of 3 + 4i )

$\longrightarrow \: \sf \: x = \dfrac{(4 - 5 \imath)(3 - 4 \imath)}{(3 + 4 \imath)(3 - 4 \imath)} \\ \\ \longrightarrow \: \sf \: x = \dfrac{12 - 31 \imath + 20 \imath {}^{2} }{ {3}^{2} - (4 { \imath})^{2} }$

Here,

• i² = - 1

• i = √-1

Now,

$\longrightarrow \: \sf \: x = \dfrac{(12 - 20) - 31 \imath}{9 + 16} \\ \\ \longrightarrow \: \boxed{ \boxed{\sf \: x = - \frac{1}{25} (8 + 31 \imath)}}$

The general form of a complex number is :

$\sf \: z = a + b\imath$

Thus,

• $\sf \: a = - \dfrac{8}{25}$
• $\sf \: b = - \dfrac{31 \imath}{25}$

The correct option is (A)

Answer 2

some important concept related to this question.....

I=√-1,,{so i^2=(-1)^( 2×1/2)=-1}

i^4=(i^2)^2=(-1)^2=1

now,lets move on question......

multiplicative inverse of (3+4i)/(4-5i)=(4-5i)/(3+4i)

now rationalize the denominator

(4-5i)(3-4i)/(3-4i)(3+4i)

=12-16i-15i-(5i×4i)/9-(4i)^2

=12+-31i-(-20)/{9-(-16)}

=12-31i-20/25

=-8-31i/25

=-8/25+ (-31/25)I

comparing it with

x+iy

we get

x=-8/25

y=-31/25

so coordinates are(-8/25,-31/25)

{hope it helps}