2. Complex Numbers and Algebra of Complex Numbers

Complex Numbers and Algebra of Complex Numbers

Example 1: Solve the complex number √-4?

Solution: √-4 = √(-1 * 4) = 2√-1 = 2i

Example 2: Solve y²+1 = 0

Solution: y²+1 = 0            

y² = -1

y = √-1 = i

Check:

L.H.S = y²+1 = i² + 1 = -1 + 1 = 0 (R.H.S)

Example 3: Find the value of √-3 + 1.

Solution: √-3 + 1

= √-1 √3 + 1

i√3 + 1

Example 4: Solve x²+16 = 0

Solution:  x²+16 = 0

x²= -16

x =√-16

x = i √16

x = i4

Check: 

L.H.S = x²+16 = (i4)² + 16 = i²4² + 16 = -16 +16 = 0 (R.H.S)

Example 5 : Simplify.

(2i)(−6i)(−7i)(2i)(−6i)(−7i)

Rewrite the expression grouping the real and imaginary parts.

(2i)(−6i)(−7i)=(2)⋅(−6)⋅(−7)⋅i⋅i⋅i

=84⋅i⋅i⋅i(2i)(−6i)(−7i)

=(2)⋅(−6)⋅(−7)⋅i⋅i⋅i=84⋅i⋅i⋅i

Simplify the imaginary part using the property of multiplying powers.

=84i³=84i²i¹=84i3=84i2i1

Recall the definition of i.

Since i²=−1:

=84(−1)i=−84i=84(−1)i=−84i

Example 6:Simplify the principal square root of √−64

Ans:

√−64

Taking the square root and substituting √-1= i

=(√−1). √64=√(−1)⋅ √64=i√64

=8i