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.
Answer: Euler.
Q3: Evaluate i⁹ + i¹⁹
Answer: i⁹ + i¹⁹
= (i²)⁴.i + (i²)⁹.i
= (-1)⁴.i + (-1)⁹.i
= i - i = 0
Q4: Evaluate i⁻³⁹
Answer: i⁻³⁹ = (i²)⁻¹⁹. i⁻¹
= (-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
Answer:
(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
Answer:
= $(4-3i)^{-1}$
= $= \frac{1}{4-3i} \times \frac{4 + 3i}{4 + 3i}$
= $\frac{4 + 3i}{4^2 - (3i)^2}$
= $\frac{4 + 3i}{16 - 9i^2}$
= $\frac{4 + 3i}{16 - 9(-1)}$
= $\frac{4 + 3i}{25}$
= $\frac{4}{25} + \frac{3}{25}i$
👉See Also:
Special Mathematical Constants
Ch2: Relations and Functions (1 Mark Q & A) Part-1
Ch2: Cartesian Product of Two Sets (Important Points)
Ch2: Relations - Domain, Range and Co-Domain (Solved Problems)
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