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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.
For example, let z = 2 + 5i be a complex number. The Real part of z is 2 and the imaginary part is 5i.
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.
Some operations on Complex numbers
(a + ib) + (c + id) = (a + c) + i(b + d).
(a - x) + i(b - y)
(a + ib).(c + id) = (ac - bd) + i(ad + bc)
(a + ib) = a - ib
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