Students will quickly understand the basics of the concept “Complex Numbers” with the aid of the complex numbers class 11 questions and answers. These questions can be used by students to briefly summarize the concept and we have also provided some practice problems so that they can comprehend the concept better. You can also get in-depth explanations for each question, and cross validate your answer as well. To learn more about complex numbers class 11, click here.
Complex Numbers Class 11 Questions with Solutions
| What are Complex Numbers?
The combination of a real number and an imaginary number is a complex number. A complex number is typically indicated by the symbol z and has the form ‘a + ib’. Both a and b are real numbers in this case. Re(z) is used to represent the value “a” as the real part, and Im(z) is used to represent the value “b” as the imaginary part . Meanwhile, “ib” is also referred to as an imaginary number. The square root of negative values can be calculated using complex numbers. Numerous fields of science, including signal processing, fluid dynamics, electromagnetism and quantum physics use complex numbers. |
1. Write the complex number i9 + i19 in the form of a + ib.
Solutions:
Given number: i9 + i19.
The expression i9 + i19 can be represented as follows:
i9 + i19 = (i2)4. i + (i2)9. i …(1)
We know that, i2 = 1.
On substituting i2 = -1, we get
i9 + i19 = (-1)4.i + (-1)9.i
i9 + i19 = 1.i + (-1).i
i9 + i19 = i – i
i9 + i19 = 0.
Therefore, i9 + i19 in the form of a + ib is 0 + i0.
2. Simplify the expression: i30 + i40 + i60.
Solutions:
Given expression: i30 + i40 + i60.
The given expression can be simplified as follows:
i30 + i40 + i60 = (i4)7. i2 + (i4)10 + (i4)15.
We know that the value of i4 is 1.
i30 + i40 + i60 = (1)7. i2 + (1)10 + (1)15.
i30 + i40 + i60 = (1)i2 + 1 + 1
i30 + i40 + i60 = -1 + 1 + 1 [since i2 = 1]
i30 + i40 + i60 = 1
Therefore, the simplification of i30 + i40 + i60 is 1.
3. Express the given expression (1 + i) (1 + 2i) in the form a + ib and find the values of a and b.
Solutions:
Given expression: (1 + i) (1 + 2i)
Hence, (1 + i) (1 + 2i) = 1(1) + 1(2i) + i + 2i(i)
(1 + i) (1 + 2i) = 1 + 2i + i + 2i2
(1 + i) (1 + 2i) = 1 + 2i + i + 2(-1) [As, i2 = -1]
(1 + i) (1 + 2i) = 1 + 2i + i – 2
(1 + i) (1 + 2i) = -1 + 3i
Hence, the expression (1 + i) (1 + 2i) in the form of a + bi is -1 + 3i.
Thus, the value of a = -1 and b = 3.
4. Solve the equation: 2x2 + x + 1 = 0
Solutions:
Given equation: 2x2 + x + 1 = 0 …(1)
Now, compare the given equation with the standard quadratic equation ax2 + bx + c = 0 …(b)
On comparing the equations (1) and (2), we get
a = 2, b = 1 and c = 1.
As we know discriminant, D = b2 – 4ac …(3)
Substitute the values a, b and c in equation (3), we get
D = (1)2 – 4(2)(1)
D = 1 – 8
D = -7
We know that the quadratic equation formula is:
Now, the above equation can be written as
Since, √-1 = i,
Hence, the solutions to the given quadratic equation are:
5. Determine the multiplicative inverse of 4 – 3i.
Solutions:
Let z = 4 – 3i.
The conjugate of 4 – 3i is 4 + 3i.
As we know, the multiplicative inverse of z is 1/z.
Hence, 1/z = 1/ (4+3i)
Therefore, the multiplicative inverse of 4 – 3i is:
As, i2 = -1
Therefore, the multiplicative inverse of 4 – 3i is (4 + 3i)/25.
6. If z1 and z2 are the two complex numbers, then show that Re (z1z2) = Re z1 Re z2 – Im z1 Im z2.
Solutions:
Let z1 = a1 + ib1 and z2 = a2+ ib2
Now, take the product of these two complex numbers.
z1z2 = (a1 + ib1)(a2+ ib2)
z1z2 = a1(a2+ ib2) + ib1(a2+ ib2)
z1z2 = a1a2 + ia1b2 + ia2b1 + i2b1b2
z1z2 = a1a2 + ia1b2 + ia2b1 – b1b2 [since i2 = -1]
Now, the above equation can be rearranged as follows:
z1z2 = a1a2 – b1b2 + ia1b2 + ia2b1
z1z2 = (a1a2 – b1b2) + i(a1b2 + a2b1)
Here, the real part is:
Re (z1z2) = a1a2 – b1b2
Thus,
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2.
Hence, Re (z1z2) = Re z1 Re z2 – Im z1 Im z2 is proved.
Also, read: Imaginary Numbers.
7. Determine the modulus and arguments of 1 – i. Also, find its polar form.
Solutions:
Given expression: 1 – i.
The complex number z = a + ib in polar form is given as z = |z| (cos θ + i sin θ) …(1)
Let z = 1 – i
Finding Modulus of 1 – i:
|z| = √(a2 + b2)
|z| = √(12 + (-1)2)
|z| = √(1+1)
|z| = √2.
Hence, the modulus of 1 – i is √2.
Finding Argument of 1 – i:
θ = arg (z) = tan-1 (|b| / |a|)
Here, |b| = 1 and |a| = 1.
Hence, θ = arg (z) = tan-1 (1/1)
θ = arg (z) = tan-1 1
Hence, θ = -π/4 radian [ As x > 0, y < 0, the complex number lies in fourth quadrant]
Now, substitute the values in (1), we get
Polar form, z = √2 (cos (-π/4) + i sin (-π/4))
z = √2 (cos (π/4) – i sin (π/4))
Therefore, the polar form of 1 – i is equal to √2 (cos (π/4) – i sin (π/4)).
8. Compute the value of (1-i)n [1- (1/i)]n for a positive integer “n”.
Solutions:
Given: (1 – i)n [1- (1/i)]n
We know that i4 = 1,
Hence, (1 – i)n [1- (1/i)]n = (1 – i)n [1- (i4/i)]n
(1 – i)n [1- (1/i)]n = (1 – i)n [1 – i3]n
Further, the above equation is written as follows:
(1 – i)n [1- (1/i)]n = (1 – i)n (1 + i)n [Since i3 = -i]
(1 – i)n [1- (1/i)]n = [(1 – i) (1 + i)]n
The expression (1 – i) (1 + i) is of the form (a – b)(a + b), which is equal to a2 – b2.
Here, a = 1, b = 1
Thus, (1 – i)n [1- (1/i)]n = (1 – i2)n
(1 – i)n [1- (1/i)]n = ( 1 – (-1))n [As, i2 = -1]
(1 – i)n [1- (1/i)]n = (1 + 1)n
(1 – i)n [1- (1/i)]n = 2n.
Therefore, the value of (1 – i)n [1- (1/i)]n is 2n.
9. Solve the given equation: |z| = z + 1 + 2i.
Solutions:
Given equation: |z| = z + 1 + 2i.
Let z = a + ib
Hence, the given equation becomes:
|a + ib| = (a + ib) + 1 + 2i
√(a2 + b2) = (a + 1) + i (b + 2) …(1)
Now, compare the real and imaginary parts,
Real part: √(a2 + b2) = a + 1 …(2)
Imaginary part: 0 = b + 2
Hence, b = -2 …(3)
Now, substitute b = -2 in (2), we get
√(a2 + (-2)2) = a + 1
Now, take squares on both sides, we get
a2 + 4 = (a+ 1)2
a2 + 4 = a2 + 1 + 2a
4 = 1 + 2a
2a = 4 – 1
2a = 3
Hence, a = 3/2 …(4)
Hence, the complex number, z = (3/2) -2i.
10. Show that arg (z1/z4) + arg (z2/z3) = 0, if z1, z2 and z3 and z4 are the two pairs of conjugate complex numbers.
Solutions:
Given that,
As we know, arg (z1/z2) = arg (z1) – arg (z2)
Hence, (z1/z4) + arg (z2/z3) = arg (z1) – arg (z4) + arg (z2) – arg (z3)
arg (z1/z4) + arg (z2/z3) = arg (z1) + arg (z2) – arg (z4) – arg (z3)
arg (z1/z4) + arg (z2/z3) = [arg (z1) + arg (z2)] – [arg (z4) + arg (z3)]
Using (1) and (2), we can write
arg (z1/z4) + arg (z2/z3) = 0 – 0
arg (z1/z4) + arg (z2/z3) = 0
Hence, arg (z1/z4) + arg (z2/z3) = 0 is proved.
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Practice Questions
Solve the following complex numbers class 11 questions.
- Write the expression i-39 in the form of a + ib.
- Solve the equation: √2x2 + x + √2 = 0.
- Write the given complex number in the polar form: (1 + 2i) / (1 – 3i).
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