Complex Numbers: Assignment

Q.1. Write the complex number  in the form x + iy.

where we have used the fact that i2 = -1, i3 = -i and i4 = 1.

Q.2. Find the three cube roots of -i. Express your answer in Cartesian form (i.e in the form x + iy)

Given z3 = -i Then z = (-i)1/3

The magnitude of -i is 1 and the principal argument is  hence

Where k = 0 ,1, 2

For, k = 1

Q.3. If z1 = 2 + i, z2 = 3 - 2i and  evaluate each of the following
(i) |3z1 - 4z2|

(i) |3z1 - 4z2| = |3(2 + i) - 4(3 - 2i) = |6 + 3i -12 + 8i| = |-6 + 11i|

Q.4. Find the fifth roots of -1- i.

Let z = (-1 -i)1/5
The modulus of -1- i is √2 and the argument is

k = 0 ,1, 2 , 3 , 4

Q.5. If (√3 + i)10 = a + ib, then find the values of a and b

Thus a = 29 and b = -29√3

Q.6. Find the value of (-2√3 - 2i)1/4 and locate the roots graphically.

k = 0 ,1, 2 , 3

These four values are generated graphically in the figure below

Q.7. If 2 + √3 is a root of quadratic equation x2 + ax + b = 0, when a, b, ∈ R, Then find the values of a and b.

Since complex roots always occur in pairs, therefore if 2 + i√3 is a root then 2 - i√3 is also a root.
Thus, sum of roots = -a = 4 ⇒ a = -4 product of roots = b = 7
⇒ (2 + i√3)(2 - i √3) = 4 + 3 = 7 = b
thus a = -4 and b = 7.

Q.8. If n = 2 , 3 , 4 ...... prove that

Consider the equation zn - 1 = 0 , whose solutions are the nth roots of unity.
1,e2πi/n , e4πi/n , e6πi/n ..... e2(n - 1)πi/n
We know that the sum of n roots of unity is 0. Hence

Q.9.

Then find the values of x and y.

Q.10. Prove that the roots of the equation 8x3 - 4x2 - 4x + 1 = 0 are cos π/7, cos 3π/7 and cos 5π/7 and hence deduce that

Let, y = cos θ + i sin θ, where θ has either of the values

Then, y7 = (cos 7θ + i sin 7θ) = -1

i.e (y + 1) (y6 - y5 + y4 - y3 + y2 - y + 1) = 0

Now the root of y = -1 corresponds to θ = π.

The roots of the equation y6 - y5 + y4 - y3 + y2 - y + 1 = 0   ....(i)
are therefore cosθ + i sinθ , where θ has either of the values

On dividing equation (1) by y3 we obtain

⇒ 8x3 - 6x - (4x2 - 2) + 2x - 1 = 0
⇒ 8x3 - 4x2 - 4x + 1 = 0

The roots of this equation are

Thus we have

Q.11.

Then find the values of x and y.

Applying C2 → C+ 3iC3 we obtain

Thus x = 0 and y = 0.
Where we have used the fact that the value of a determinant is zero if any of its row or column is zero.

Q.12. Solve the equation z2 + (2i - 3) z + 5 - i = 0

The roots of a quadratic equation az2 + bz + c = 0

Let us calculate the square roots of (-15 - 8i)
-15 - 8i = 17 [cos(θ + 2kπ) + i sin (θ + 2kπ)] where k = 0 and 1.
Where, cos θ = -15/17 and sin θ = -8/17

since θ is an angle in the third quadrant, θ/2 is an angle in the second quadrant.
Hence  and the two values of (-15 - 8i)1/2 are -1 + 4i

and 1 - 4i.
Hence we can write

Q.13. If 5z2/7z1 is purely imaginary, then find the value of  where z1 ≠ 0

5z2/7z1 is purely imaginary ⇒ 5z2/7z1 = ki, where k ∈ R

Q.14. Prove that for m = 2, 3, ...

The roots of zm = are z = 1, e2πi/m , e4πi/m , e6πi/m ... e2(m-1)πi/m

Then we can write  zm - 1 = (z - 1)(z - e2πi/m) (z - e4πi/m)...(z - e2(m - 1)πi/m)
Divid ing both sides by (z - 1) and then letting z = 1(realizing that
We find m = (1 - e2πi/m) (1 - e4πi/m)...(1 - e2(m - 1)πi/m)  ....(i)
Taking the complex conjugate of both sides of equation (I) yields
m = (1 - e2πi/m) (1 - e4πi/m)...(1 - e2(m - 1)πi/m)  ....(ii)
Multiplying (I) by (II) using

We obtain,

Q.15. If the equation |z-i| / |z+i| = 3 represent  a circle in the complex plane, then find the radius and the centre of the circle.

⇒ x2 + (y -1)2 = 9[x2 + (y + 1)2]
⇒ x2 + y2 - 2y + 1 = 9x2 + 9y2 + 18y + 9
⇒ 8x2 + 8y2 + 20y + 8 = 0

this is the equation of a circle with center at  and of radius

Q.16. If |z| < 1, Then 1 + z + z2 + z3 + ... 1/(1-z). Using this fact and assuming that |a| < 1, prove the following
(a) 1 + a cosθ + a2 cos2θ + a3 cos3θ ....

(b) a sinθ + a2 sin2θ + a3 sin3θ + ..

Let z = ae , then since |a| < 1 we can write

⇒ (1 + a cosθ + a2 cos 2θ + ....) + i (a sinθ + a2 sin 2θ + ....)

⇒ (1 + a cosθ + a2 cos2θ +...) + i(a sinθ + a2 sin 2θ + ....)

Equating the real and imaginary parts we obtain

Q.17. Let z1 and z2 be two complex numbers such that z1 + z2 and z1z2 are both real then find the value of  denotes the complex conjugate z2.

Let z1 = a+ ib and z2 = c + id, then z1 + z2 is real ⇒ (a + c) + i (b + d) is real
⇒ b + d = 0 ⇒ d = -b
z1z2 in real ⇒ (ac - bd) + i (ad + bc) in real. ⇒ ad + bc = 0 ⇒ a (-b) + bc = 0 ⇒ a = c

Q.18. Find the real and Imaginary parts of
(a) e3iz
(b) cos 2z
(c) z2e2z
(d) sin h2z
(e) z coshz

(a) e3iz = e3i(x+iy) = e3ix-3y ⇒ e3iz = e-3y. e3ix = e-3y [cos 3x + i sin3x]

Thus the real and imaging parts are u(x, y) = e-3y cos 3x and v(x, y) = e-3y sin 3x.
(b) cos 2 z = cos (2 x+ 2iy)

= cos 2x · cos (2iy) - sin 2x sin (2iy)
⇒ cos 2z = cos 2x cosh 2y - i sin 2x sinh 2y

Thus, u (x, y) = cos 2x cosh 2y

And, v (x, y) = - sin 2x sinh 2y
(c) z2e2z = (x + iy)2 e2(x + iy)

⇒ z2 e2z = (x2 + 2ixy - y2) e2x·e2iy
= z2 e2z = e2x [(x2 - y2)cos 2y - 2xy sin 2y] +ie2x[2xy cos 2y + (x2 - y2) sin 2y]

[cos 2y · sinh 2x + i sin 2y cosh 2x]

u(x, y) = cos 2y sinh 2x , v (x, y) = sin 2y cosh 2x
(e)

Hence, u (x, y) = cos 2x · cosh 2y and v (x, y) = sin 2x sinh 2y

= (x cosh x cos y - y sinh x sin y) + i (y cosh x cos y + x sinh x sin y)
Thus u (x, y) = cosh x cos y - y sinh x sin y and v (x, y) = y cosh x cos y + x sinh x sin y

Q.19. Find the locus of points representing the complex numbers z for which |z| - 2 = |z - i| - |z + 5i| = 0

|z| - 2 = |z - i| - |z + 5i| = 0
⇒ |z| = 2 and |z - i| = |z + 5i|
⇒ z lies on a circle |z| = 2 and the perpendicular bisector of the  line segment joining (0, -5) and (0, 1) that is y = -2 putting y = -2 in |z| = 2
i.e x2 + y2 = 4 we obtain x = 0
Hence the locus of z is a single point (0, -2).

Q.20. Obtain all the values of
(a) In(√3 - i)
(b) ln (3i)
(c) ln (-√2- i√2)
(d) (1 - √2i)i
And then find the principal value in each case.

(a) ln (√3 - i)

And the principal argument is

Where n = 0, ±1, ±2, ±3 ......
The principal value of (3 - i) is obtained when n = 0, Thus the principal value is

(b) The complex number z = 3i has modulus 3 and the principal argument π/2

The principal values of ln (3i) occurs when n = 0, Thus the principal value

(c) The complex number  has the modulus  and the principal argument

Hence the general values of

Where n = 0, ±1, ±2, ±3,......
The principal values of is obtained when n = 0, thus the principa l values is

(d) First we find the value of

Where  k = 0,1

when,

when,  we obtain

when  the principal value is

when,  the principal value is

Q.21. If (1+x)/(1-x) = cos2θ + i sin2θ, then find the value of α such that x = α tan θ.

Using componendo and dividendo we can write

Hence, the value of α is i.

Q.22. Express each of the following complex numbers in polar and Eulerian form (using the principal argument -π < arg z ≤ π).
(i) 2 + 2√3i
(ii)
(iii) (√2 + √2i)12
(iv) (9 + 9i)3

(i) we have

Since θ lies in the first quadrant θ = α
Hence,   θ = π/α,
Hence the polar form is
Eulerian form is z = eiπ/3.
(ii) Given

Rationalising the denominator gives

Thus z lies on the negative x - axis and its principle argument is θ = π

Hence, r = 3 and θ = π
Hence the polar form is z = 3[cos(π) + i sin (π)]. The Eulerian form is z = 3e

(iii) Let z = (√2 + √2i)
Then,

Since θ lies in the first quadrant hence θ = α = π/4.

Thus, in polar form

and the Eulerian form of z is z = 2eiπ/4

Using Euler’s formula
(cosθ + i sinθ)n =  cos nθ + i sin nθ
We obtain
z12 = 212 [cos 3π + i sin 3π]
In terms of principle angle we can write

z12 = 212 (cos π + i sin π)
The eulerian form of z12 is
z12 = 212ei(3π) = 212 e

(iv) Let z = 9 + 9i

Since θ lies in the first quadrant, hence θ = α = π/4.

Using the Euler’s formula, (cos θ + isin θ)n = cos nθ + i sin nθ

We can write ;

Since   is the principal argument.
The Eulerian form of

Q.23.  then find arg (z) (where -π < argz ≤ π).

Since z = 1 lies on the real axis hence its principal argument is 0.

Q.24.  where x and y are reals then find the values of x and y.

The Three roots of unity are

Hence we can write -(-ω)50 = (x + iy)
Since ω3 = 1 hence we can write -ω2 = (x + iy)

The document Complex Numbers: Assignment | Mathematical Methods - Physics is a part of the Physics Course Mathematical Methods.
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## Mathematical Methods

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## FAQs on Complex Numbers: Assignment - Mathematical Methods - Physics

 1. What is the importance of complex numbers in the IIT JAM exam?
Ans. Complex numbers play a crucial role in the IIT JAM exam, particularly in subjects like Mathematics and Physics. Many problems and concepts in these subjects involve complex numbers, such as complex algebra, complex analysis, and complex functions. Having a strong understanding of complex numbers is essential to solve complex equations, analyze electrical circuits, and comprehend quantum mechanics, among other topics frequently encountered in the IIT JAM exam.
 2. How can complex numbers be represented geometrically?
Ans. Complex numbers can be represented geometrically using the Argand plane or the complex plane. In this representation, the real part of a complex number represents the x-coordinate, while the imaginary part represents the y-coordinate. The complex number itself can be represented as a point in the complex plane. The magnitude of the complex number corresponds to its distance from the origin, and the angle it forms with the positive real axis represents its argument or phase.
 3. What are the properties of complex conjugates?
Ans. Complex conjugates are pairs of complex numbers having the same real part but opposite imaginary parts. The properties of complex conjugates include: - The product of a complex number and its conjugate is equal to the square of its magnitude. - The sum of a complex number and its conjugate is equal to twice the real part of the complex number. - The difference between a complex number and its conjugate is equal to twice the imaginary part of the complex number. These properties are frequently used to simplify complex expressions and equations in the IIT JAM exam.
 4. How are complex numbers used in solving polynomial equations?
Ans. Complex numbers are used extensively in solving polynomial equations, particularly those with complex roots. The Fundamental Theorem of Algebra states that any polynomial equation of degree 'n' has 'n' complex roots (including repeated roots). These complex roots can be found using methods like synthetic division, factoring, and the quadratic formula. By using complex numbers, we can accurately determine the roots of polynomials, even when they involve complex solutions.
 5. How can complex numbers be used to solve problems in electrical circuits?
Ans. Complex numbers are widely used in analyzing electrical circuits due to their ability to represent both magnitude and phase. By using complex numbers, we can represent voltages, currents, and impedance in a compact and efficient manner. The concept of phasors, which are complex numbers representing sinusoidal functions, allows us to analyze circuits in the frequency domain. This approach simplifies calculations involving alternating current (AC) circuits, resonance, impedance matching, and complex power analysis, all of which are important topics in the IIT JAM exam.

## Mathematical Methods

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