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Test: Introduction To Complex Numbers - Question 1

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 1

Detailed Solution for Test: Introduction To Complex Numbers - Question 2

(3-4i) / (2-3i)*(2+3i) / (2+3i) = (6+9i-8i+12) / 13=(18/13)+(i/13)

Test: Introduction To Complex Numbers - Question 3

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 3

(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

Test: Introduction To Complex Numbers - Question 4

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 4

Given: (2−√3i)(2+√3i) + 2 − 4i

(2−√3i)(2+√3i) = 7

⇒ 7 + 2 - 4i

⇒ 9 - 4i

Test: Introduction To Complex Numbers - Question 5

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 5

-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

Test: Introduction To Complex Numbers - Question 6

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 6

(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

Detailed Solution for Test: Introduction To Complex Numbers - Question 7

(i^{-997}) = 1/(i^{997}), 1/((i^{4})^{249}) × i

Since (i^{4}) = 1, (i^{4}) / i = (i^{3})

= - i (Since i^{2} = -1 , therefore, i^{3} = - i)

Test: Introduction To Complex Numbers - Question 8

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 8

(8 - 4i) - (-2 - 3i) + (-10 + 3i)

=> 8 - 4i + 2 + 3i-10 + 3i

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

Test: Introduction To Complex Numbers - Question 9

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 9

(2-3i)^{2} = 4 + 9 (i)^{2} - 2.2.3i

= 4 - 9 - 12i since, i^{2} = -1

= - 5 - 12 i

Detailed Solution for Test: Introduction To Complex Numbers - Question 10

Test: Introduction To Complex Numbers - Question 11

Express the following in standard form : i^{20} + (1 - 2i)^{3}

Detailed Solution for Test: Introduction To Complex Numbers - Question 11

Given, i^{20} + (1 - 2i)^{3}

We knoe that i = √-1

i^{2} = -1

Now put the values in given equation

= i^{20} + (1 - 2i)^{3}

= ( i^{2})^{10} + { 1 - 8i^{3} - 6i + 12i^{2 }}

= 1 +1 - 8i^{3} - 6i + 12i^{2}

=1 +1 - 8i^{2}.i^{1} - 6i + 12i^{2}

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

= -10 + 2i

Detailed Solution for Test: Introduction To Complex Numbers - Question 12

(-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)

Test: Introduction To Complex Numbers - Question 13

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

Detailed Solution for Test: Introduction To Complex Numbers - Question 13

This is called conjugate of complex no.

z = a+ib. conjugate of z = a-ib

- sign is put before i

Detailed Solution for Test: Introduction To Complex Numbers - Question 14

Detailed Solution for Test: Introduction To Complex Numbers - Question 15

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

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