Test: Introduction To Complex Numbers - Commerce MCQ

first write above equation in complex number format , ie using iota
(3-4i) / (2-3i)*(2+3i) / (2+3i) = (6+9i-8i+12) / 13=(18/13)+(i/13)

(x + iy)(3 + 2i) = (1 + i)
x + iy = (1 + i)/(3 + 2i)
x + iy = [(1 + i) * (3 - 2i)] / [(3 + 2i)*(3 - 2i)]
x + iy = (3 + 3i - 2i + 2) / [(3)² + (2)²]
x + iy = (5 + i)/[ 9 + 4]
= (5 + i) / 13
=> 13x + 13iy = 5+i
13x = 5         13y = 1
x = 5/13         y = 1/13

Given: (2−√3i)(2+√3i) + 2 − 4i​

(2−√3i)(2+√3i) = 7

⇒ 7 + 2 - 4i

⇒ 9 - 4i

-2 + 5i
multiplicative inverse of -2 + 5i is
1/(-2+5i)
= 1/(-2+5i) * ((-2-5i)/(-2-5i))
= -2-5i/(-2)^2 -(5i)^2
= -2-5i/4-(-25)
= -2-5i/4+25
= -2-5i/29
= -2/29 -5i/29

(x + iy) (3 + 2i)
= 3x + 2xi + 3iy + 3i*y = 1+i
= 3x-2y + i(2x+3y) = 1+i
= 3x-2y-1 = 0 ; 2x + 3y -1 = 0  
on equating real and imaginary parts on both sides
on solving two equations
x= 5/13 ; y = 1/13  

(i^-997) = 1/(i⁹⁹⁷), 1/((i⁴)²⁴⁹) × i

Since (i⁴) = 1, (i⁴) / i = (i³)

= - i  (Since i² = -1 , therefore, i³ = - i)

(8 - 4i) - (-2 - 3i) + (-10 + 3i)
=> 8 - 4i + 2 + 3i-10 + 3i
=> 8 + 2 - 10 -  4i + 3i + 3i  =>0 + 2i

(2-3i)² = 4 + 9 (i)² - 2.2.3i
= 4 - 9 - 12i     since, i² = -1
 = - 5 - 12 i 

Given, i²⁰ + (1 - 2i)³

We knoe that i = √-1

i² = -1

Now put the values in given equation

= i²⁰ + (1 - 2i)³

= ( i²)¹⁰ + { 1 - 8i³ - 6i + 12i² }

= 1 +1 - 8i³ - 6i + 12i²

=1 +1 - 8i².i¹ - 6i + 12i²

=1 + 1 + 8i - 6i -12

=  -10 + 2i

(-i)(3i) +2(-i) =-3(i^2)-2i =-3(-1)-2i =3-2i since i=√-1 =3+(-2)i comparing with a+bi,we get b=(-2)

This is called conjugate of complex no. 
z = a+ib. conjugate of z = a-ib
- sign is put before i

Complete answers is in 3 steps:
1. Conjugate = 3+4i
2. Modulus = √3^2 + 4^2 =5
3. Multiplicative inverse = conjugate/square of modulus = 3+4i/5^2 = 3+4i/25