How to find cube of an imaginary number?

We can all find the square root of real positive numbers but it was not possible with negative real numbers. Therefore to find the roots of negative numbers the concept of complex numbers come into play. Complex numbers are the numbers that are combination of real and imaginary numbers. They are expressed in the form x + iy where x and y are real numbers and i is the iota. x is the real part while iy is the imaginary part of the complex number. 

• Real numbers are those which can be positive, negative, zero, rational, irrational etc and can be plotted on number line. 
• Imaginary numbers are those numbers which cannot be plotted on number line and they are represented in the form of ‘xi’ where i is the iota and x is the real number. 

For example, let z = 2 + 5i be a complex number. The Real part of z is 2 and the imaginary part is 5i.

What happens when you cube an imaginary number? 

Answer:

Since imaginary numbers are of the form ‘xi’ where x is the real number and i is iota. So when an imaginary number is cubed the product always gives a negative result.

When “i”, the imaginary number is squared, the answered obtained is -1, 

i = √(-1)

i² = -1

Now, in order to obtain cube of the imaginary number, multiply with i again,

i × i² = -i

i³ = -i

More about Iota

The value of iota is √-1. To differentiate the real part from imaginary part we use iota. The role of iota comes into play when we need to find square root of a negative number. 

1. The value of i² = -1
2. The value of i³ = -i
3. The value of i⁴ = 1

Some operations on Complex numbers

• Addition: When two complex numbers, say, a + ib and x + iy are added, the real parts are added separately and imaginary parts are added separately that is, 

(a + ib) + (c + id) = (a + c) + i(b + d).

• Subtraction: Just like addition, subtraction also follows the same rule. Let there be two complex numbers a + ib and x + iy. The result is, 

(a – x) + i(b – y)

• Multiplication: When two complex numbers are multiplied the real parts are multiplied together, then real part with imaginary part and then imaginary parts. 

(a + ib).(c + id) = (ac – bd) + i(ad + bc)

• Conjugate: means inverse of sign. If the operator is positive then it will be converted to negative and vice versa. 

(a + ib) = a – ib

• Division: The division is performed by multiplying the numerator and denominator by the conjugate of denominator.

Sample Problems

Question 1: Find the square of √5i.

Solution:

(√5i)² = √5i × √5i 

= -5i

Question 2: Find the value of (-8i)³

Solution: 

(-8i)³ = -8 × -8 × -8 × i × i × i 

= -512 × -i

= 512i

Question 3: Find the value of (a + ib)²

Solution: 

Expanding we get,

(a + ib)² = a² + 2aib – b²

Question 4: Simplify √-56.

Solution: 

56 can be expressed in the form of 7 × 2³

Therefore √-56 = √-(7 × 2³) 

= (√7 × 2³) × (√-1) 

= 2√14 i

Question 5: Find the square of (-1 – 2i).

Solution: 

(-1 – 2i)² = [-(1 + 2i)]²

= (1)² + 2 × (1) × (2i) + (2i)²

= 1 – 4 + 4i 

= -3 + 4i

Question 6: Find the cube of 5i.

Solution: 

(5i)³ = 5 × 5 × 5 × i × i × i 

= 125 × (-i) 

= -125i

Question 7: Find the value of (a – ib)²

Solution: 

(a – ib)² = (a)² – 2 × a × (bi) + (bi)²

= a² – 2iab -b²