Complex Numbers and Quadratic Equation Test

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Complex Numbers and Quadratic Equation Test - 1

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Weekly Quiz Competition

Question 1
2 / -0.83

Find the result in the form a + ib of (2-√-25) / (1+√-16)

A

B

C

D

Solution
Question 2
2 / -0.83

Express the following in standard form :

A

B

C

D

Solution

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)
Question 3
2 / -0.83

Find the real numbers x and y such that : (x + iy)(3 + 2i) = 1 + i

A
x = 5/13 , y = 1/13

B
x = 5/3 , y = 1/3

C

D
x = 15/13 , y = 1/13

Solution

(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
Question 4
2 / -0.83

Express the following in standard form : (2 –√3i) (2 + √3i) + 2 –4i

A
3 + 4i

B
4i

C
9 - 4i

D
13 + 4i

Solution

Given: (2 −√3i)(2+√3i) + 2 −4i
(2 −√3i)(2+√3i) = 7
⇒7 + 2 - 4i
⇒9 - 4i
Question 5
2 / -0.83

Find the reciprocal (or multiplicative inverse) of -2 + 5i

A

B

C

D

Solution

-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
Question 6
2 / -0.83

Find the real numbers x and y such that : (x + iy)(3+2i) = 1 + i

A

B

C

D

Solution

(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
Question 7
2 / -0.83

Write in the simplest form: (i)^-997

A
- i

B
1

C
-1

D
i

Solution

(i^-997 ) = 1/(i⁹⁹⁷ ), 1/((i⁴ )²⁴⁹ ) ×i
Since (i⁴ ) = 1, (i⁴ ) / i = (i³ )
= - i (Since i² = -1 , therefore, i³ = - i)
Question 8
2 / -0.83

Express the following in standard form : (8 - 4i) - (-2 - 3i) + (-10 + 3i)

A
3+3i

B
2 + 2i

C
1 + 2i

D
0 + 2i

Solution

(8 - 4i) - (-2 - 3i) + (-10 + 3i)
=>8 - 4i + 2 + 3i-10 + 3i
=>8 + 2 - 10 - 4i + 3i + 3i =>0 + 2i
Question 9
2 / -0.83

Express the following in standard form : (2-3i)²

A
5 + 2i

B
8 + 12i

C
-5 –12i

D
8 + 2i

Solution

(2-3i)² = 4 + 9 (i)² - 2.2.3i
= 4 - 9 - 12i since, i² = -1
= - 5 - 12 i
Question 10
2 / -0.83

Find the multiplicative inverse of 2 −3i

B
2+3i

C
−2 −3i

D
3i

Solution
Question 11
2 / -0.83

Express the following in standard form : i²⁰ + (1 - 2i)³

A
-10 + 2i

B
-10 + 4i

C
-10 + i

D
-1 + 2i

Solution

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
Question 12
2 / -0.83

Imaginary part of −i(3i + 2) is:

A
−2

B
2

C
3

D
−3

Solution

(-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)
Question 13
2 / -0.83

For a complex number a+ib, a-ib is called its:

A
Opposite

B
Conjugate

C
Reciprocal

D
Additive inverse

Solution

This is called conjugate of complex no.
z = a+ib. conjugate of z = a-ib
- sign is put before i
Question 14
2 / -0.83

Express the following in standard form :

A
3 + 4i

B
13 + 4i

C
6 + 4i

D
4i

Solution
Question 15
2 / -0.83

The multiplicative inverse of 3 –4i is:

A
(3 + 4i)/25

B
3 + 4i

C
-3 + 4i/5

D
-3 + 4i

Solution

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