Complex Numbers: Assignment | Mathematical Methods - Physics PDF Download

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,e^2πi/n , e^4πi/n , e^6πi/n ..... e^{2(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 → C₂ + 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 z^m = are z = 1, e^2πi/m , e^4πi/m , e^6πi/m ... e^2(m-1)πi/m 

Then we can write  z^m - 1 = (z - 1)(z - e^2πi/m) (z - e^4πi/m)...(z - e^{2(m - 1)πi/m})
Divid ing both sides by (z - 1) and then letting z = 1(realizing that
We find m = (1 - e^2πi/m) (1 - e^4πi/m)...(1 - e^{2(m - 1)πi/m})  ....(i)
Taking the complex conjugate of both sides of equation (I) yields
m = (1 - e^2πi/m) (1 - e^4πi/m)...(1 - e^{2(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^iθ , 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) e^3iz
(b) cos 2z
(c) z²e^2z
(d) sin h2z 
(e) z coshz

(a) e^3iz = e^3i(x+iy) = e^3ix-3y ⇒ e^3iz = e^-3y. e^3ix = 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) z²e^2z = (x + iy)² e^{2(x + iy)}

⇒ z² e^2z = (x² + 2ixy - y²) e^2x·e^2iy 
= z² e^2z = e^2x [(x² - y²)cos 2y - 2xy sin 2y] +ie^2x[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 = e^iπ/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^iπ  

(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 = 2e^iπ/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 = 2¹²e^i(3π) = 2¹² e^iπ 

(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)