# Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.1

**Question 1: Evaluate the following:**

**(i) i ^{457}**

**(ii) i ^{528}**

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**(iii) 1/i ^{58}**

**(iv) i ^{37} + 1/i^{67}**

**(v) [i ^{41} + 1/i^{257}]^{9}**

**(vi) (i ^{77} + i^{70} + i^{87} + i^{414})^{3}**

**(vii) i ^{30} + i^{40} + i^{60}**

**(viii) i ^{49} + i^{68} + i^{89} + i^{110}**

**Solution:**

We know that i = √-1

i

^{2}= -1i

^{3}= -ii

^{4}= 1

(i) i^{457}To find i

^{n},As n is greater than 4, so we divide 457 by 4, we get

When dividing 457 by 4 we get quotient (p) as 114 and remainder (q) as 1

Therefore, substituting the value of p and q in the equation i

^{n}= i^{4p+q }we get.i

^{457}= i^{4(114) + 1}i

^{457}= i^{4(114)}× ii

^{457}= (1)^{114}× i [As i^{4}= 1, therefore 1^{114}= 1]i

^{457}= i

(ii) i^{528}To find i

^{n},As n is greater than 4, so we divide 528 by 4, we get

When dividing 528 by 4 we get quotient (p) as 132 and the remainder (q) as 0

Therefore, substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get.i

^{528}= i^{4(132)}i

^{528}= (1)^{132}[As i^{4}= 1, therefore 1^{132}= 1]i

^{528}= 1

(iii) 1/ i^{58}To find i

^{n},As n is greater than 4, so we divide 58 by 4, we get

When dividing 58 by 4 we get quotient (p) as 14 and the remainder (q) as 2

Therefore, substituting the value of p and q in the equation i

^{n}= i4p+q we get.1/ i

^{58}= 1/ i^{4(14) + 2}1/ i

^{58}= 1/ i^{4(14)}× i^{2 }[As i^{4}= 1, therefore 1^{14}= 1]^{ }1/ i

^{58}= 1/ i^{2}[since, i^{2}= -1]1/ i

^{58 }= 1/-11/ i

^{58}= -1

(iv) i^{37}+ 1/i^{67}To find i

^{n},As n is greater than 4, so we divide 37 and 67 by 4, we get

When dividing 37 by 4 we get quotient (p) as 9 and the remainder (q) as 1

When dividing 67 by 4 we get quotient (p) as 16 and the remainder (q) as 3

Therefore, substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get.i

^{37}+ 1/i^{67}= i^{4(9)+1}+ 1/ i^{4(16)+3}= i

^{4(9)}×i + 1/ i^{4(16)}×i^{3}= i + 1/i

^{3 }[As, i^{4}= 1]Multiplying numarator and denominator by

i,we get= i + i/i

^{4}= i + i

i

^{37}+ 1/i^{67}= 2i

(v) [i^{41}+ 1/i^{257}]^{9}To find i

^{n},As n is greater than 4, so we divide 41 and 257 by 4, we get

When dividing by 4 we get quotient (p) as 10 and the remainder (q) as 1

When dividing 257 by 4 we get quotient (p) as 64 and the remainder (q) as 1

Therefore, substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,[i

^{41}+ 1/i^{257}] = [i^{4(10)+1}+ 1/ i^{4(64)+1}]^{9}= [ i

^{4(10)}×i + 1/ i^{4(64)}×i ]^{9}= [i + 1/i]

^{9}[As, i^{4}= 1 and 1/i = -1]= [i – i]

^{9}= 0

(vi) (i^{77}+ i^{70}+ i^{87}+ i^{414})^{3}To find i

^{n},As n is greater than 4, so we divide 77, 70, 87 and 414 by 4, we get

When dividing 77 by 4 we get quotient (p) as 19 and the remainder (q) as 1.

When dividing 70 by 4 we get quotient (p) as 17 and the remainder (q) as 2.

When dividing 87 by 4 we get quotient (p) as 21 and the remainder (q) as 3.

When dividing 414 by 4 we get quotient (p) as 103 and the remainder (q) as 2 .

Therefore, substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,(i

^{77}+ i^{70}+ i^{87}+ i^{414})^{3}= (i^{4(17)+ 1}+ i^{4(21) + 2}+ i^{4(21) + 3}+ i^{4(103) + 2})^{3}= (i

^{4(17)}× i + i^{4(21)}× i^{2}+ i^{4(21)}× i^{3}+ i^{4(103)}× i^{2})^{3 }[As, i^{4}= 1 ]= (i + i

^{2}+ i^{3}+ i^{2})^{3}[As, i^{3}= – i, i^{2}= – 1]= (i + (– 1) + (– i) + (– 1))

^{3}= (– 2)

^{3}= – 8

(vii) i^{30}+ i^{40}+ i^{60}To find i

^{n},As n is greater than 4, so we divide 30, 40 and 60 by 4, we get

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

When dividing 40 by 4 we get quotient (p) as 10 and the remainder (q) as 0.

When dividing 60 by 4 we get quotient (p) as 15 and the remainder (q) as 0.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,i

^{30}+ i^{40}+ i^{60}= i^{4(7) + 2}+ i^{4(10)}+ i^{4(15)}= i

^{4(7)}× i^{2}+ i^{4(10)}+ i^{4(15)}= i

^{2}+ 1^{10}+ 1^{15 }[As, i^{4}= 1 and i^{2}= -1]= – 1 + 1 + 1

= 1

(viii) i^{49}+ i^{68}+ i^{89}+ i^{110}To find i

^{n},As n is greater than 4, so we divide 49, 68, 89 and 110 by 4, we get

When dividing 49 by 4 we get quotient (p) as 12 and the remainder (q) as 1.

When dividing 68 by 4 we get quotient (p) as 17 and the remainder (q) as 0.

When dividing 89 by 4 we get quotient (p) as 22 and the remainder (q) as 1.

When dividing 110 by 4 we get quotient (p) as 27 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,i

^{49}+ i^{68}+ i^{89}+ i^{110}= i^{4(12) + 1}+ i^{4(17)}+ i^{4(22) + 1}+ i^{4(27) + 2}= i

^{4(12)}× i + i^{4(17)}+ i^{4(22)}× i + i^{4(27)}× i^{2}= i + 1 + i – 1 [As, i4 = 1]

= 2i

**Question 2: Show that 1 + i**^{10} + i^{20} + i^{30} is a real number?

^{10}+ i

^{20}+ i

^{30}is a real number?

**Solution:**

Given: 1 + i

^{10}+ i^{20}+ i^{30}To find i

^{n},As n is greater than 4, so we divide 10, 20, and 30 by 4, we get

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 2.

When dividing 20 by 4 we get quotient (p) as 5 and the remainder (q) as 0.

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,1 + i

^{10}+ i^{20}+ i^{30 }= 1 + i^{4(2) + 2 }+ i^{4(5)}+ i^{4(7) + 2}= 1 + i

^{4(2)}× i^{2}+ i^{4(5)}+ i^{4(7)}× i^{2}= 1 – 1 + 1 – 1 [As, i

^{4}= 1, i^{2}= – 1]= 0

Therefore , 1 + i

^{10}+ i^{20 }+ i^{30}=0, and 0 is a real number.

**Question 3: Find the values of the following expressions:**

**(i) i ^{49} + i^{68} + i^{89} + i^{110}**

**(ii) i ^{30} + i^{80} + i^{120}**

**(iii) i + i ^{2} + i^{3} + i^{4}**

**(iv) i ^{5} + i^{10 }+ i^{15}**

**(v) [i ^{592} + i^{590} + i^{588} + i^{586} + i^{584}]/[i^{582} + i^{580} + i^{578} + i^{576} + i^{574}]**

**(vi) 1 + i ^{2} + i^{4} + i^{6} + i^{8} + … + i^{20}**

**(vii) (1 + i) ^{6} + (1 – i)^{3}**

**Solution:**

(i) i^{49}+ i^{68}+ i^{89}+ i^{110}To find i

^{n},As n is greater than 4, so we divide 49, 68, 89 and 110 by 4, we get

When dividing 49 by 4 we get quotient (p) as 12 and the remainder (q) as 1.

When dividing 68 by 4 we get quotient (p) as 17 and the remainder (q) as 0

When dividing 89 by 4 we get quotient (p) as 22 and the remainder (q) as 1.

When dividing 110 by 4 we get quotient (p) as 27 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,i

^{49}+ i^{68}+ i^{89}+ i^{110}= i^{4(12) + 1}+ i^{4(17)}+ i^{4(22) + 1}+ i^{4(27) + 2}= i

^{4(12)}× i + i^{4(17)}+ i^{4(22)}× i + i^{4(27)}× i^{2}= i + 1 + i – 1 [As, i4 = 1]

= 2i

Therefore, i

^{49}+ i^{68}+ i^{89}+ i^{110}= 2i

(ii) i^{30}+ i^{80 }+ i^{120}To find i

^{n},As n is greater than 4, so we divide 30, 80, and 120 by 4, we get

When dividing 30 by 4 we get quotient (p) as 7 and the remainder (q) as 2.

When dividing 80 by 4 we get quotient (p) as 20 and the remainder (q) as 0.

When dividing 120 by 4 we get quotient (p) as 30 and the remainder (q) as 0.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,i

^{30}+ i^{80}+ i^{120}= i4(7) + 2 + i4(20) + i4(30)= i

^{4(7)}× i^{2}+ i^{4(20)}+ i^{4(30)}= – 1 + 1 + 1 [As, i

^{4}= 1, i^{2}= – 1]= 1

Therefore, i

^{30}+ i^{80}+ i^{120}= 1

(iii) i + i^{2}+ i^{3}+ i^{4}i + i

^{2}+ i^{3}+ i^{4}= i + i^{2}+ i^{2+1}+ i^{4}= i + i

^{2}+ i^{2}×i + i4= i – 1 + (– 1) × i + 1 [As, i

^{4}= 1, i^{2}= – 1]= i – 1 – i + 1

= 0

Therefore, i + i

^{2}+ i^{3}+ i^{4}= 0

(iv) i^{5}+ i^{10}+ i^{15}To find i

^{n},As n is greater than 4, so we divide 5, 10, and 10 by 4, we get

When dividing 5 by 4 we get quotient (p) as 1 and the remainder (q) as 1.

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 1.

When dividing 15 by 4 we get quotient (p) as 3 and the remainder (q) as 3.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,i

^{5}+ i^{10}+ i^{15}= i^{4(1) + 1}+ i^{4(2) + 2}+ i^{4(3) + 3}= i

^{4}×i + i^{4(2)}×i^{2}+ i^{4(3)}×i^{3}= i

^{4}×i + i^{4(2)}×i^{2}+ i^{4(3)}×i^{2}×i [As, i^{4}= 1, i^{2}= – 1]= 1×i + 1 × (– 1) + 1 × (– 1)×i

= i – 1 – i

= – 1

Therefore, i

^{5}+ i^{10}+ i^{15}= -1

(v) [i^{592}+ i^{590}+ i^{588}+ i^{586}+ i^{584}] / [i^{582 }+ i^{580}+ i^{578}+ i^{576}+ i^{574}][i

^{592}+ i^{590}+ i^{588}+ i^{586}+ i^{584}] / [i^{582}+ i^{580}+ i^{578}+ i^{576}+ i^{574}]= [i

^{10}(i^{582}+ i^{580}+ i^{578}+ i^{576}+ i^{574}) / (i^{582}+ i^{580}+ i^{578}+ i^{576}+ i^{574})] [Taking i^{10}as common from numerator]= i

^{10}To find i

^{n},As n is greater than 4, so

When dividing 10 by 4 we get quotient (p) as 2 and the remainder (q) as 2.

Therefore substituting the value of p and q in the equation i

^{n}= i^{4p+q}we get,= i

^{4(2)+2}= i

^{4(2) }× i^{2}= -1 [As, i

^{4}= 1, i^{2}= -1]= -1

Therefore, [i

^{592}+ i^{590}+ i^{588}+ i^{586}+ i^{584}] / [i^{582}+ i^{580}+ i^{578}+ i^{576}+ i^{574}] = -1

(vi) 1 + i^{2}+ i^{4}+ i^{6}+ i^{8}+ … + i^{20}When n is greater than 4, so we divide n by 4,

Here we will divide all the values greater than 4 i.e 6, 8, 10, 12, 14, 16, 18, and 20.

1 + i

^{2}+ i^{4}+ i^{6}+ i^{8}+ … + i^{20}= 1 + i^{2}+ i^{4}+ i^{4+2}+ i^{4+4 }+ … + i^{4(5)}= 1 + (– 1) + 1 + (– 1) + 1 + … + 1 [As, i4 = 1, i2 = -1]

= 1

Therefore, 1 + i

^{2}+ i^{4}+ i^{6}+ i^{8}+ … + i^{20}= 1

(vii) (1 + i)^{6}+ (1 – i)^{3}(1 + i)

^{6}+ (1 – i)^{3}= [(1 + i)^{2}]^{3}+ (1 – i)^{2}(1 – i)= [1 + i

^{2}+ 2i]^{3}+ (1 + i^{2}– 2i)(1 – i) [By using formula (a+b)^{2}= a^{2}+ b^{2}+ 2ab ]= [1 – 1 + 2i]

^{3}+ (1 – 1 – 2i)(1 – i)= (2i)

^{3}+ (– 2i)(1 – i)= 8i

^{3}+ (– 2i) + 2i^{2}= – 8i – 2i – 2 [As, i

^{3}= – i, i^{2}= – 1]= – 10i – 2

= – 2 – 10i

Therefore (1 + i)

^{6 }+ (1 – i)^{3}= – 2 – 10i