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Q.1. Write the complex number Complex Numbers: Assignment | Mathematical Methods - Physics in the form x + iy.

Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics

where we have used the fact that i2 = -1, i3 = -i and i4 = 1.
Complex Numbers: Assignment | Mathematical Methods - Physics


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 Complex Numbers: Assignment | Mathematical Methods - Physics hence
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Where k = 0 ,1, 2
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
For, k = 1 

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics


Q.3. If z1 = 2 + i, z2 = 3 - 2i and Complex Numbers: Assignment | Mathematical Methods - Physics evaluate each of the following 
(i) |3z1 - 4z2|
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics

(i) |3z1 - 4z2| = |3(2 + i) - 4(3 - 2i) = |6 + 3i -12 + 8i| = |-6 + 11i|
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics


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 Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics

k = 0 ,1, 2 , 3 , 4
Complex Numbers: Assignment | Mathematical Methods - Physics


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

Complex Numbers: Assignment | Mathematical Methods - Physics
Thus a = 29 and b = -29√3


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

Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics

k = 0 ,1, 2 , 3
Complex Numbers: Assignment | Mathematical Methods - Physics

These four values are generated graphically in the figure below
Complex Numbers: Assignment | Mathematical Methods - Physics


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
Complex Numbers: Assignment | Mathematical Methods - Physics

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
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics


Q.9. 
Complex Numbers: Assignment | Mathematical Methods - Physics

Then find the values of x and y.

Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics


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
Complex Numbers: Assignment | Mathematical Methods - Physics

 Let, y = cos θ + i sin θ, where θ has either of the values
Complex Numbers: Assignment | Mathematical Methods - Physics
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

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
On dividing equation (1) by y3 we obtain
Complex Numbers: Assignment | Mathematical Methods - Physics
⇒ 8x3 - 6x - (4x2 - 2) + 2x - 1 = 0
⇒ 8x3 - 4x2 - 4x + 1 = 0 

The roots of this equation are

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Thus we have
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics


Q.11.
Complex Numbers: Assignment | Mathematical Methods - Physics
Then find the values of x and y.

Applying C2 → C+ 3iC3 we obtain
Complex Numbers: Assignment | Mathematical Methods - Physics
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
Complex Numbers: Assignment | Mathematical Methods - Physics
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
Complex Numbers: Assignment | Mathematical Methods - Physics
since θ is an angle in the third quadrant, θ/2 is an angle in the second quadrant.
Hence Complex Numbers: Assignment | Mathematical Methods - Physics and the two values of (-15 - 8i)1/2 are -1 + 4i

and 1 - 4i.
Hence we can write
Complex Numbers: Assignment | Mathematical Methods - Physics


Q.13. If 5z2/7z1 is purely imaginary, then find the value of Complex Numbers: Assignment | Mathematical Methods - Physics where z1 ≠ 0

5z2/7z1 is purely imaginary ⇒ 5z2/7z1 = ki, where k ∈ R
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics


Q.14. Prove that for m = 2, 3, ... 
Complex Numbers: Assignment | Mathematical Methods - Physics

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 Complex Numbers: Assignment | Mathematical Methods - Physics
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
Complex Numbers: Assignment | Mathematical Methods - Physics

We obtain,
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics


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.

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
⇒ x2 + (y -1)2 = 9[x2 + (y + 1)2]
⇒ x2 + y2 - 2y + 1 = 9x2 + 9y2 + 18y + 9
⇒ 8x2 + 8y2 + 20y + 8 = 0
Complex Numbers: Assignment | Mathematical Methods - Physics
this is the equation of a circle with center at Complex Numbers: Assignment | Mathematical Methods - Physics and of radius
Complex Numbers: Assignment | Mathematical Methods - Physics


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 ....
Complex Numbers: Assignment | Mathematical Methods - Physics
(b) a sinθ + a2 sin2θ + a3 sin3θ + ..
Complex Numbers: Assignment | Mathematical Methods - Physics

Let z = ae , then since |a| < 1 we can write
Complex Numbers: Assignment | Mathematical Methods - Physics
⇒ (1 + a cosθ + a2 cos 2θ + ....) + i (a sinθ + a2 sin 2θ + ....)
Complex Numbers: Assignment | Mathematical Methods - Physics
⇒ (1 + a cosθ + a2 cos2θ +...) + i(a sinθ + a2 sin 2θ + ....)
Complex Numbers: Assignment | Mathematical Methods - Physics
Equating the real and imaginary parts we obtain
Complex Numbers: Assignment | Mathematical Methods - Physics


Q.17. Let z1 and z2 be two complex numbers such that z1 + z2 and z1z2 are both real then find the value of Complex Numbers: Assignment | Mathematical Methods - Physics 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
Complex Numbers: Assignment | Mathematical Methods - Physics


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)

Complex Numbers: Assignment | Mathematical Methods - Physics
⇒ 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]
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
[cos 2y · sinh 2x + i sin 2y cosh 2x]

u(x, y) = cos 2y sinh 2x , v (x, y) = sin 2y cosh 2x
(e)
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
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

Complex Numbers: Assignment | Mathematical Methods - Physics

|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)

Complex Numbers: Assignment | Mathematical Methods - Physics

And the principal argument is Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
Where n = 0, ±1, ±2, ±3 ......
The principal value of (3 - i) is obtained when n = 0, Thus the principal value is
Complex Numbers: Assignment | Mathematical Methods - Physics
(b) The complex number z = 3i has modulus 3 and the principal argument π/2
Complex Numbers: Assignment | Mathematical Methods - Physics

The principal values of ln (3i) occurs when n = 0, Thus the principal value
Complex Numbers: Assignment | Mathematical Methods - Physics
(c) The complex number Complex Numbers: Assignment | Mathematical Methods - Physics has the modulus Complex Numbers: Assignment | Mathematical Methods - Physics and the principal argument
Complex Numbers: Assignment | Mathematical Methods - Physics
Hence the general values of
Complex Numbers: Assignment | Mathematical Methods - Physics
Where n = 0, ±1, ±2, ±3,......
The principal values ofComplex Numbers: Assignment | Mathematical Methods - Physics is obtained when n = 0, thus the principa l values is

Complex Numbers: Assignment | Mathematical Methods - Physics

(d) First we find the value of Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
Where  k = 0,1
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
when, Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
when, Complex Numbers: Assignment | Mathematical Methods - Physics we obtain

Complex Numbers: Assignment | Mathematical Methods - Physics
when Complex Numbers: Assignment | Mathematical Methods - Physics the principal value isComplex Numbers: Assignment | Mathematical Methods - Physics

when, Complex Numbers: Assignment | Mathematical Methods - Physics the principal value is
Complex Numbers: Assignment | Mathematical Methods - Physics


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

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
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) Complex Numbers: Assignment | Mathematical Methods - Physics
(iii) (√2 + √2i)12 
(iv) (9 + 9i)3 

(i) we have 

Complex Numbers: Assignment | Mathematical Methods - Physics
Since θ lies in the first quadrant θ = α
Hence,   θ = π/α, Complex Numbers: Assignment | Mathematical Methods - Physics
Hence the polar form is Complex Numbers: Assignment | Mathematical Methods - Physics
Eulerian form is z = eiπ/3.
(ii) Given Complex Numbers: Assignment | Mathematical Methods - Physics

Rationalising the denominator gives
Complex Numbers: Assignment | Mathematical Methods - Physics

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,
Complex Numbers: Assignment | Mathematical Methods - Physics
Since θ lies in the first quadrant hence θ = α = π/4.
Complex Numbers: Assignment | Mathematical Methods - Physics

Thus, in polar form
Complex Numbers: Assignment | Mathematical Methods - Physics
and the Eulerian form of z is z = 2eiπ/4 
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics
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
Complex Numbers: Assignment | Mathematical Methods - Physics

Since θ lies in the first quadrant, hence θ = α = π/4.
Complex Numbers: Assignment | Mathematical Methods - Physics

Complex Numbers: Assignment | Mathematical Methods - Physics

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

We can write ;
Complex Numbers: Assignment | Mathematical Methods - Physics
Since Complex Numbers: Assignment | Mathematical Methods - Physics  is the principal argument.  
The Eulerian form of Complex Numbers: Assignment | Mathematical Methods - Physics


Q.23. Complex Numbers: Assignment | Mathematical Methods - Physics then find arg (z) (where -π < argz ≤ π).

Complex Numbers: Assignment | Mathematical Methods - Physics

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


Q.24. Complex Numbers: Assignment | Mathematical Methods - Physics where x and y are reals then find the values of x and y.

Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
The Three roots of unity are  
Complex Numbers: Assignment | Mathematical Methods - Physics
Complex Numbers: Assignment | Mathematical Methods - Physics
Hence we can write -(-ω)50 = (x + iy)
Since ω3 = 1 hence we can write -ω2 = (x + iy)
Complex Numbers: Assignment | Mathematical Methods - Physics

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