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Q.1. Write the complex number in the form x + iy.
where we have used the fact that i^{2} = 1, i^{3} = i and i^{4} = 1.
Q.2. Find the three cube roots of i. Express your answer in Cartesian form (i.e in the form x + iy)
Given z^{3} = 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 z_{1} = 2 + i, z_{2} = 3  2i and evaluate each of the following
(i) 3z_{1}  4z_{2}
(i) 3z_{1}  4z_{2} = 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 isk = 0 ,1, 2 , 3 , 4
Q.5. If (√3 + i)^{10 }= a + ib, then find the values of a and b
Thus a = 2^{9} and b = 2^{9}√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 x^{2} + 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 z^{n}  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 8x^{3}  4x^{2}  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, y^{7} = (cos 7θ + i sin 7θ) = 1i.e (y + 1) (y^{6}  y^{5} + y^{4}  y^{3} + y^{2}  y + 1) = 0
Now the root of y = 1 corresponds to θ = π.
The roots of the equation y^{6}  y^{5} + y^{4}  y^{3} + y^{2}  y + 1 = 0 ....(i)
are therefore cosθ + i sinθ , where θ has either of the values
On dividing equation (1) by y^{3} we obtain
⇒ 8x^{3}  6x  (4x^{2}  2) + 2x  1 = 0
⇒ 8x^{3}  4x^{2}  4x + 1 = 0The roots of this equation are
Thus we have
Q.11.
Then find the values of x and y.
Applying C_{2} → C_{2 }+ 3iC_{3} 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 z^{2} + (2i  3) z + 5  i = 0
The roots of a quadratic equation az^{2} + 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 + 4iand 1  4i.
Hence we can write
Q.13. If 5z_{2}/7z_{1} is purely imaginary, then find the value of where z_{1} ≠ 0
5z_{2}/7z_{1} is purely imaginary ⇒ 5z_{2}/7z_{1} = 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(m1)πi/m}
Then we can write z^{m}  1 = (z  1)(z  e^{2}^{πi/m}) (z  e^{4}^{πi/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^{2}^{πi/m}) (1  e^{4}^{πi/m})...(1  e^{2(m  1)}^{πi/m}) ....(i)
Taking the complex conjugate of both sides of equation (I) yields
m = (1  e^{2}^{πi/m}) (1  e^{4}^{πi/m})...(1  e^{2(m  1)}^{πi/m}) ....(ii)
Multiplying (I) by (II) usingWe obtain,
Q.15. If the equation zi / z+i = 3 represent a circle in the complex plane, then find the radius and the centre of the circle.
⇒ x^{2} + (y 1)^{2} = 9[x^{2} + (y + 1)^{2}]
⇒ x^{2} + y^{2}  2y + 1 = 9x^{2} + 9y^{2} + 18y + 9
⇒ 8x^{2} + 8y^{2} + 20y + 8 = 0
this is the equation of a circle with center at and of radius
Q.16. If z < 1, Then 1 + z + z^{2} + z^{3} + ... 1/(1z). Using this fact and assuming that a < 1, prove the following
(a) 1 + a cosθ + a^{2} cos2θ + a^{3} cos^{3}θ ....
(b) a sinθ + a^{2} sin2θ + a^{3} sin3θ + ..
Let z = ae^{iθ} , then since a < 1 we can write
⇒ (1 + a cosθ + a^{2} cos 2θ + ....) + i (a sinθ + a^{2} sin 2θ + ....)
⇒ (1 + a cosθ + a^{2} cos2θ +...) + i(a sinθ + a^{2} sin 2θ + ....)
Equating the real and imaginary parts we obtain
Q.17. Let z_{1} and z_{2} be two complex numbers such that z_{1} + z_{2} and z_{1}z_{2} are both real then find the value of denotes the complex conjugate z_{2}.
Let z_{1} = a+ ib and z_{2} = c + id, then z_{1} + z_{2} is real ⇒ (a + c) + i (b + d) is real
⇒ b + d = 0 ⇒ d = b
z_{1}z_{2} 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) z^{2}e^{2z}
(d) sin h2z
(e) z coshz
(a) e^{3iz} = e^{3i(x+iy)} = e^{3ix3y} ⇒ 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 2yThus, u (x, y) = cos 2x cosh 2y
And, v (x, y) =  sin 2x sinh 2y
(c) z^{2}e^{2z }= (x + iy)^{2} e^{2(x + iy)}
⇒ z^{2} e^{2z }= (x^{2} + 2ixy  y^{2}) e^{2x}·e^{2iy}
= z^{2} e^{2z} = e^{2x} [(x^{2}  y^{2})cos 2y  2xy sin 2y] +ie^{2x}[2xy cos 2y + (x^{2}  y^{2}) 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 x^{2} + y^{2} = 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/2The 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 iswhen, the principal value is
Q.21. If (1+x)/(1x) = 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) GivenRationalising 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
z^{12} = 2^{12 }[cos 3π + i sin 3π]
In terms of principle angle we can writez^{12} = 2^{12} (cos π + i sin π)
The eulerian form of z^{12} is
z^{12} = 2^{12}e^{i(3π)} = 2^{12} 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 ω^{3} = 1 hence we can write ω^{2} = (x + iy)
18 docs24 tests
