# Class 11 - Maths - Complex Numbers - Solved Problems - Part 1

Q1: What are imaginary numbers?

Answer: If the square of a given number is negative then such a number is called an imaginary number.

Q2: Name the mathematician who was the first to introduce the symbol i (iota) for square root of -1.

Q3: Evaluate i⁹ + i¹⁹

= (i²)⁴.i + (i²)⁹.i

= (-1)⁴.i + (-1)⁹.i

= i - i = 0

Q4: Evaluate i⁻³⁹

= (-1)⁻¹⁹. i⁻¹

= $\frac{1}{(-1)^{19}} \times \frac{1}{i}$

= $\frac{1}{(-1)} \times \frac{1}{i}$

= $\frac{-1}{1} \times \frac{1}{i}$

= $\frac{i^2}{1} \times \frac{1}{i}$

= i

Q5: Simplify $4\sqrt{-4} + 5\sqrt{-9} -3\sqrt{-16}$

Answer: $4\sqrt{-4} + 5\sqrt{-9} -3\sqrt{-16}$

= $\because \sqrt{-4} = 2i, \sqrt{-9} = 3i \: and\: \sqrt{-16} = 4i$

= $(4\times 2i) + (5 \times 3i) - (3 \times 4i)$

= 8i + 15i - 12i

= 11i

Q6: What are complex numbers?

Answer: The numbers of the form (a + ib), where a and b are real numbers and i = $\sqrt{}-1$, are known as complex numbers.

For a complex number, z = a + ib,

we have a = real part of z,

and b = imaginary part of z,

Q7: Express the complex number in the form of a + ib

(a) 3(7+i7)+i(7+i7)

(b) (1-i) - (-1+i6)

(c) (1 -i)⁴

(a) 3(7+i7)+i(7+i7)

= 21 + 21i + 7i + 7i²

= 21 + 28i + 7(-1)

= 14 + 28i

(b) (1-i) - (-1+i6)

= (1 - i) + 1 - 6i)

= 1+1 -i - 6i

= 2 - 7i

(c) (1 -i)⁴

= [(1 -i)²]²

= [1 + i² -2i]²

= [1 - 1 -2i]²

= [-2i]²

= 4i²

= -4

Q8: Find the multiplicative inverse of 4 - 3i

= $(4-3i)^{-1}$

= $= \frac{1}{4-3i} \times \frac{4 + 3i}{4 + 3i}$

= $\frac{4 + 3i}{4^2 - (3i)^2}$

= $\frac{4 + 3i}{16 - 9i^2}$

= $\frac{4 + 3i}{16 - 9(-1)}$

= $\frac{4 + 3i}{25}$

= $\frac{4}{25} + \frac{3}{25}i$