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Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.2 | Set 2

  • Last Updated : 30 Apr, 2021

Question 14. If\left(\frac{1-i}{1+i}\right)^{100}=a+ib , find (a, b).

Solution:

We have,

=>\left(\frac{1-i}{1+i}\right)^{100}=a+ib

=>\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]^{100}=a+ib

=>\left(\frac{1+i^2-2i}{1-i^2}\right)^{100}=a+ib



=>\left(\frac{-2i}{2}\right)^{100}=a+ib

=> (−i)100 = a + ib

=> a + ib = 1

On comparing real and imaginary parts on both sides, we get,

=> (a, b) = (1, 0)

Question 15. If a = cos θ + i sin θ, find the value of\frac{1+a}{1-a} .

Solution:

Given a = cos θ + i sin θ, we get,

\frac{1+a}{1-a} =\frac{1+cosθ+isinθ}{1-cosθ-isinθ}



=\frac{(1+cosθ+isinθ)(1-cosθ+isinθ)}{(1-cosθ-isinθ)(1-cosθ+isinθ)}

=\frac{(1+isinθ)^2-cos^2θ}{(1-cosθ)^2-i^2sin^2θ}

=\frac{1+i^2sin^2θ+2isinθ-cos^2θ}{1+cos^2θ-2cosθ-i^2sin^2θ}

=\frac{1+2isinθ-sin^2θ-cos^2θ}{1-2cosθ+cos^2θ+sin^2θ}

=\frac{1+2isinθ-1}{1-2cosθ+1}

=\frac{2isinθ}{2-2cosθ}

=\frac{isinθ}{1-cosθ}

=\frac{2isin\frac{θ}{2}cos\frac{θ}{2}}{2sin^2\frac{θ}{2}}

=icot\frac{θ}{2}

Therefore, the value of\frac{1+a}{1-a} isicot\frac{θ}{2} .



Question 16. Evaluate the following :

(i) 2x3 + 2x2 − 7x + 72, when x = (3−5i)/2

Solution:

We have, x = (3−5i)/2

=> 2x = 3 − 5i

=> 2x − 3 = −5i

=> (2x − 3)2 = 25i2

=> 4x2 + 9 − 12x = −25

=> 4x2 − 12x + 34 = 0

=> 2x2 − 6x + 17 = 0

Now, 2x3 + 2x2 − 7x + 72 = x (2x2 − 6x + 17) + 6x2 − 17x + 2x2 − 7x + 72

= x (0) + 8x2 − 24x + 72

= 4 (2x2 − 6x + 17) + 4

= 4 (0) + 4

= 4

Therefore, the value of 2x3 + 2x2 − 7x + 72 is 4.

(ii) x4 − 4x3 + 4x2 +8x +44, when x = 3 + 2i

Solution:

We have, x = 3 + 2i

=> x − 3 = 2i

=> (x − 3)2 = (2i)2

=> x2 + 9 − 6x = 4i2

=> x2 − 6x + 9 + 4 = 0

=> x2 − 6x + 13 = 0

Now, x4 − 4x3 + 4x2 + 8x + 44 = x2 (x2 − 6x + 13) + 6x3 − 13x2 − 4x3 + 4x2 + 8x + 44

= 2x3 − 9x2 + 8x + 44

= 2x (x2 − 6x + 13) + 12x2 − 26x − 9x2 + 8x + 44

= 3x2 − 18x + 44

= 3 (x2 − 6x + 13) + 5

= 5

Therefore, the value of x4 − 4x3 + 4x2 + 8x + 44 is 5.

(iii) x4 + 4x3 + 6x2 + 4x + 9, when x = −1 + i√2

Solution:

We have, x = −1 + i√2



=> x + 1 = i√2

=> (x + 1)2 = 2i2

=> x2 + 1 + 2x = −2

=> x2 + 2x + 3 = 0

Now, x4 + 4x3 + 6x2 + 4x + 9 = x2 (x2 + 2x + 3) − 2x3 − 3x2 + 4x3 + 6x2 + 4x + 9

= 2x3 + 3x2 + 4x + 9

= 2x (x2 + 2x + 3) − 4x2 − 6x + 3x2 + 4x + 9

= − x2 − 2x + 9

= − (x2 + 2x + 3) + 3 + 9

= 3 + 9

= 12

Therefore, the value of x4 + 4x3 + 6x2 + 4x + 9 is 12.

(iv) x6 + x4 + x2 + 1, when x = (1+i)/√2

Solution:

We have, x = (1+i)/√2

=> √2x = 1 + i

=> 2x2 = 1 + i2 + 2i

=> 2x2 = 2i

=> 4x4 = 4i2

=> x4 = −1

=> x4 + 1 = 0

Now, x6 + x4 + x2 + 1 = (x6 + x2) + (x4 +1)

= x6 + x2

= x2 (x4 + 1)

= 0

Therefore, the value of x6 + x4 + x2 + 1 is 0.

(v) 2x4 + 5x3 + 7x2 − x + 41, when x = −2 − √3i

Solution:

We have, x = −2 − √3i

x2 = (−2 − √3i)2 = 4 + 4√3i + 3i2 = 1 + 4√3i

x3 = (1 + 4√3i) (−2 − √3i) = −2 − √3i − 8√3i −12i2 = 10 − 9√3i

x4 = (1 + 4√3i)2 = 1 + 8√3i + 48i2 = −47 + 8√3i



Now, 2x4 + 5x3 + 7x2 − x + 41 becomes,

= 2(−47 + 8√3i) + 5(10 − 9√3i) + 7(1 + 4√3i) − (−2 − √3i) + 41

= −94 + 16√3i + 50 − 45√3i + 7 + 28√3i + 2 + √3i + 41

= 6

Therefore, the value of 2x4 + 5x3 + 7x2 − x + 41 is 6.

Question 17. For a positive integer n, find the value of (1−i)n (1−1/i)n.

Solution:

We have,

(1−i)n (1−1/i)n = (1−i)n(\frac{i-1}{i})^n

=\left[\frac{(1-i)^2}{-i}\right]^n

=\left(\frac{1+i^2-2i}{-i}\right)^n

=\left(\frac{-2i}{-i}\right)^n

= 2n

Therefore, the value of (1−i)n (1−1/i)n is 2n.

Question 18. If (1+i)z = (1−i)\bar{z} , then show that z = −i\bar{z} .

Solution:

We have,

=> (1+i)z = (1−i)\bar{z}

=> z =(\frac{1-i}{1+i})\bar{z}

=> z =\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]\bar{z}

=> z =\left[\frac{1+i^2-2i}{1-i^2}\right]\bar{z}

=> z =\left[\frac{-2i}{2}\right]\bar{z}



=> z = −i\bar{z}

Hence proved.

Question 19. Solve the system of equations: Re(z2) = 0, |z| = 2.

Solution:

Let z = x + iy.

Now z2 = (x + iy)2

= x2 + i2y2 + 2xyi

= x2 − y2 + 2xyi

We have, Re(z2) = 0

=> x2 − y2 = 0 . . . . (1)

Also, it is given, |z| = 2.

=>\sqrt{x^2+y^2} = 2

=> x2 + y2 = 4 . . . . (2)

Solving (1) and (2), we get, x = ±√2 and y = ±√2.

Therefore, x + iy = ±√2 ± √2i .

Question 20. If\frac{z-1}{z+1} is purely imaginary number (z≠−1), find the value of |z|.

Solution:

Let z = x + iy

We have,\frac{z-1}{z+1}

=\frac{x-1+iy}{x+1+iy}

=\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}

=\frac{x^2+x-ixy-x-1+iy+ixy+iy+y^2}{(x+1)^2-(iy)^2}



=\frac{x^2+y^2-1+2iy}{x^2+y^2+1+2xy}

=\frac{x^2+y^2-1}{x^2+y^2+1+2xy}+\frac{2yi}{x^2+y^2+1+2xy}

As the complex number is purely imaginary, therefore,

=> Re(z) = 0

=>\frac{x^2+y^2-1}{x^2+y^2+1+2xy} = 0

=> x2 + y2 = 1

=>\sqrt{x^2+y^2} = 1

=> |z| = 1

Therefore, the value of |z| is 1.

Question 21. If z1 is a complex number other than −1 such that |z1| = 1 and z2 =\frac{z_1-1}{z_1+1} ,then show that the real parts of z2 is zero.

Solution:

Given |z| = 1

=> |z|2 = 1

=> x2 + y2 = 1 . . . . (1)

Let z1 = x + iy and z2 = a + ib.

According to the question, we have,

=> z2 =\frac{z_1-1}{z_1+1}

=> a + ib =\frac{x+iy-1}{x+iy+1}

=> a + ib =\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}

=> a + ib =\frac{x^2-1+y^2-iyx+iy+iyx+iy}{(x+1)^2-(iy)^2}

=> a + ib =\frac{(x^2+y^2)-1+2iy}{(x+1)^2-(iy)^2}



Using (1) we get,

=> a + ib =\frac{1-1+2iy}{(x+1)^2-(iy)^2}

=> a + ib =\frac{2iy}{(x+1)^2-(iy)^2}

On comparing the real and imaginary parts on both sides, we get a = 0.

Therefore, the real parts of z2 is 0. Hence proved.

Question 22. If |z+1| = z + 2(1+i), find z.

Solution:

Let z = x + iy. According to the question, we have,

=> |x + iy + 1| = x + iy + 2(1 + i)

=>\sqrt{(x+1)^2+y^2} = (x + 2) + i(y + 2)

On comparing the real and imaginary parts, we get

=> y + 2 = 0

=> y = −2

And also,

=> x + 2 =\sqrt{(x+1)^2+y^2}

=> (x + 2)2 = (x+1)2 + y2

=> x2 + 4 + 4x = x2 + 2x + 1+ y2

=> 2x = y2 − 3

=> 2x = 4 − 3

=> 2x = 1

=> x = 1/2

Therefore, z = x + iy = 1/2 −2i.

Question 23. Solve the equation: |z| = z + 1 + 2i.

Solution:

Let z = x + iy. According to the question, we have,

=> |z| = z + 1 + 2i

=> |x + iy| = x + iy + 1 + 2i

=>\sqrt{x^2+y^2} = (x + 1) + (y + 2)i

=> x2 + y2 = (x+1)2 + (y+2)2i2 + 2 (x+1) (y+2)i

=> x2 + y2 = x2+1 + 2x − y2 − 1 + 2y + 2 (x+1) (y+2)i

=> 2y2 − 2x + 4y + 4 = 2i (x+1) (y+2)

=> y2 − x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

=> (x+1) (y+2) = 0

=> x = −1 and y = −2

Also, y2 − x + 2y + 2 = 0

Taking x = −1, we get y2 − (−1) + 2y + 2 = 0

=> y2 + 2y + 3 = 0, which doesn’t have a solution as the roots are imaginary.

Taking y = −2, (4 − x −4 + 2) = 0

=> x = 2

Therefore, z = x + iy = 2 − 2i.

Question 24. What is the smallest positive integer n for which (1+i)2n = (1−i)2n?

Solution:



We are given,

=> (1+i)2n = (1−i)2n

=>\left(\frac{1+i}{1-i}\right)^{2n} = 1

=>\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^{2n} = 1

=>\left[\frac{1+i^2+2i}{1-i^2}\right]^{2n} = 1

=>\left[\frac{2i}{2}\right]^{2n} = 1

=> i2n = 1

=> i2n = i4

=> 2n = 4

=> n = 2

Therefore, the smallest positive integer n for which (1+i)2n = (1−i)2n is 2.

Question 25. If z1, z2, z3 are complex numbers such that |z1| = |z2| = |z3| =|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}| = 1, then find the value of |z1 + z2 + z3|.

Solution:

We are given,

|z1| = |z2| = |z3| =|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}| = 1

Now, |z1 + z2 + z3| =|\frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+\frac{z_3\bar{z_3}}{\bar{z_3}}|

=|\frac{|z_1|^2}{\bar{z_1}}+\frac{|z_2|^2}{\bar{z_2}}+\frac{|z_3|^2}{\bar{z_3}}|

=|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|

= 1

Therefore, the value of |z1 + z2 + z3| is 1.

Question 26. Find the number of solutions of z2 + |z|2 = 0.

Solution:



Let z = x + iy. We have,

=> z2 + |z|2 = 0

=> (x + iy)2 + |x + iy|2 = 0

=> x2 + i2y2 + 2xyi + x2 + y2 = 0

=> x2 − y2 + 2xyi + x2 + y2 = 0

=> 2x2 + 2xyi = 0

On comparing the real and imaginary parts on both sides, we get

=> 2x2 = 0 and 2xy = 0

=> x = 0 and y ∈ R

Therefore, z = 0 + iy, where y ∈ R.

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