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

Last Updated : 30 Apr, 2021

### Question 14. If, find (a, b).

Solution:

We have,

=>

=>

=>

=>

=> (âˆ’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 Î¸ + i sin Î¸, find the value of.

Solution:

Given a = cos Î¸ + i sin Î¸, we get,

=

=

=

=

=

=

=

=

=

=

Therefore, the value ofis.

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

Solution:

We have, x = (3âˆ’5i)/2

=> 2x = 3 âˆ’ 5i

=> 2x âˆ’ 3 = âˆ’5i

=> (2x âˆ’ 3)2 = 25i2

=> 4x2 + 9 âˆ’ 12x = âˆ’25

=> 4x2 âˆ’ 12x + 34 = 0

=> 2x2 âˆ’ 6x + 17 = 0

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

= x (0) + 8x2 âˆ’ 24x + 72

= 4 (2x2 âˆ’ 6x + 17) + 4

= 4 (0) + 4

= 4

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

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

Solution:

We have, x = 3 + 2i

=> x âˆ’ 3 = 2i

=> (x âˆ’ 3)2 = (2i)2

=> x2 + 9 âˆ’ 6x = 4i2

=> x2 âˆ’ 6x + 9 + 4 = 0

=> x2 âˆ’ 6x + 13 = 0

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

= 2x3 âˆ’ 9x2 + 8x + 44

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

= 3x2 âˆ’ 18x + 44

= 3 (x2 âˆ’ 6x + 13) + 5

= 5

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

### (iii) x4 + 4x3 + 6x2 + 4x + 9, when x = âˆ’1 + iâˆš2

Solution:

We have, x = âˆ’1 + iâˆš2

=> x + 1 = iâˆš2

=> (x + 1)2 = 2i2

=> x2 + 1 + 2x = âˆ’2

=> x2 + 2x + 3 = 0

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

= 2x3 + 3x2 + 4x + 9

= 2x (x2 + 2x + 3) âˆ’ 4x2 âˆ’ 6x + 3x2 + 4x + 9

= âˆ’ x2 âˆ’ 2x + 9

= âˆ’ (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)/âˆš2

Solution:

We have, x = (1+i)/âˆš2

=> âˆš2x = 1 + i

=> 2x2 = 1 + i2 + 2i

=> 2x2 = 2i

=> 4x4 = 4i2

=> x4 = âˆ’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 âˆ’ x + 41, when x = âˆ’2 âˆ’ âˆš3i

Solution:

We have, x = âˆ’2 âˆ’ âˆš3i

x2 = (âˆ’2 âˆ’ âˆš3i)2 = 4 + 4âˆš3i + 3i2 = 1 + 4âˆš3i

x3 = (1 + 4âˆš3i) (âˆ’2 âˆ’ âˆš3i) = âˆ’2 âˆ’ âˆš3i âˆ’ 8âˆš3i âˆ’12i2 = 10 âˆ’ 9âˆš3i

x4 = (1 + 4âˆš3i)2 = 1 + 8âˆš3i + 48i2 = âˆ’47 + 8âˆš3i

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

= 2(âˆ’47 + 8âˆš3i) + 5(10 âˆ’ 9âˆš3i) + 7(1 + 4âˆš3i) âˆ’ (âˆ’2 âˆ’ âˆš3i) + 41

= âˆ’94 + 16âˆš3i + 50 âˆ’ 45âˆš3i + 7 + 28âˆš3i + 2 + âˆš3i + 41

= 6

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

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

Solution:

We have,

(1âˆ’i)n (1âˆ’1/i)n = (1âˆ’i)n

=

=

=

= 2n

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

### Question 18. If (1+i)z = (1âˆ’i), then show that z = âˆ’i.

Solution:

We have,

=> (1+i)z = (1âˆ’i)

=> z =

=> z =

=> z =

=> z =

=> z = âˆ’i

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 âˆ’ y2 + 2xyi

We have, Re(z2) = 0

=> x2 âˆ’ y2 = 0 . . . . (1)

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

=>= 2

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

Solving (1) and (2), we get, x = Â±âˆš2 and y = Â±âˆš2.

Therefore, x + iy = Â±âˆš2 Â± âˆš2i .

### Question 20. Ifis purely imaginary number (zâ‰ âˆ’1), find the value of |z|.

Solution:

Let z = x + iy

We have,

=

=

=

=

=

As the complex number is purely imaginary, therefore,

=> Re(z) = 0

=>= 0

=> x2 + y2 = 1

=>= 1

=> |z| = 1

Therefore, the value of |z| is 1.

### Question 21. If z1 is a complex number other than âˆ’1 such that |z1| = 1 and z2 =,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 =

=> a + ib =

=> a + ib =

=> a + ib =

=> a + ib =

Using (1) we get,

=> a + ib =

=> a + ib =

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)

=>= (x + 2) + i(y + 2)

On comparing the real and imaginary parts, we get

=> y + 2 = 0

=> y = âˆ’2

And also,

=> x + 2 =

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

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

=> 2x = y2 âˆ’ 3

=> 2x = 4 âˆ’ 3

=> 2x = 1

=> x = 1/2

Therefore, z = x + iy = 1/2 âˆ’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

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

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

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

=> 2y2 âˆ’ 2x + 4y + 4 = 2i (x+1) (y+2)

=> y2 âˆ’ x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

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

=> x = âˆ’1 and y = âˆ’2

Also, y2 âˆ’ x + 2y + 2 = 0

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

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

Taking y = âˆ’2, (4 âˆ’ x âˆ’4 + 2) = 0

=> x = 2

Therefore, z = x + iy = 2 âˆ’ 2i.

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

Solution:

We are given,

=> (1+i)2n = (1âˆ’i)2n

=>= 1

=>= 1

=>= 1

=>= 1

=> i2n = 1

=> i2n = i4

=> 2n = 4

=> n = 2

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

### Question 25. If z1, z2, z3 are complex numbers such that |z1| = |z2| = |z3| == 1, then find the value of |z1 + z2 + z3|.

Solution:

We are given,

|z1| = |z2| = |z3| == 1

Now, |z1 + z2 + z3| =

=

=

= 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 âˆ’ 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 âˆˆ R

Therefore, z = 0 + iy, where y âˆˆ R.

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