StudentAidUnit 1: Complex Numbers — Exercise 1 1
10th Class Mathematics · Unit 1: Complex Numbers
1.1.1.Simplify the following: (i) i^5 (ii) i^16 (iii) (-i)^-19 (iv) 27i^-26 (v) i^11 + i^5 (vi) (i^4+i^3+i^2+i)^2 (vii) (i^8/i^5)^-5 (viii) i^13 × i^29
Given
(i)
i5
= i4+1 = i4 · i = 1 · i
Result
(i)
i5 = i
Given
(ii)
i16
= (i4)4 = 14
Result
(ii)
i16 = 1
Given
(iii)
(-i)-19
(-i)19 = (-1)19 · i19 = -i19
i19 = i16 · i3 = (i4)4 · i3 = 1 · (-i) = -i ⇒ (-i)19 = -(-i) = i
(-i)-19 = 1i = 1i × ii = ii2 = i-1
Result
(iii)
(-i)-19 = -i
Given
(iv)
27i-26
= 27 · frac{1}{i26 = 27 · 1(i4)6 · i2 = 271 · (-1)
Result
(iv)
27i-26 = -27
Given
(v)
i11 + i5
= i8+3 + i4 · i = (i4)2 · i3 + 1 · i = 1 · (-i) + i
Result
(v)
i11 + i5 = 0
Given
(vi)
(i4+i3+i2+i)2
= (1+(-i)+(-1)+i)2 = (0)2
Result
(vi)
(i4+i3+i2+i)2 = 0
Given
(vii)
left(i8i5right)-5
= (i8-5)-5 = (i3)-5 = frac{1}{i15 = 1i12 · i3 = 11 · (-i) = -1i
= -1i × ii = -ii2 = -i-1
Result
(vii)
left(i8i5right)-5 = i
Given
(viii)
i13 × i29
= i42 = (i4)10 · i2 = 110 · (-1)
Result
(viii)
i13 × i29 = -1
1.1.2.Write in terms of i: (i) 2 + √-4 (ii) 3 - √-7 (iii) 2/5 + √-16/5 (iv) √2 - √-3
Given
(i)
2+sqrt{-4}
= 2+sqrt4 · sqrt{-1} = 2+2i
Result
(i)
2+2i
Given
(ii)
3-sqrt{-7}
= 3-sqrt7 · sqrt{-1} = 3-sqrt7 i
Result
(ii)
3-sqrt7 i
Given
(iii)
25+frac{sqrt{-16}{5}
= frac25+frac{sqrt{16} · sqrt{-1}{5} = frac25+4i5
Result
(iii)
frac25+frac45 i
Given
(iv)
sqrt2-sqrt{-3}
= sqrt2-sqrt3 · sqrt{-1} = sqrt2-sqrt3 i
Result
(iv)
sqrt2-sqrt3 i
1.1.3.Find the values of x and y: (i) (2x+5)+(y-3)i = 1+2i (ii) (3x+2)-(4-y)i = 5+3i (iii) (2+i)x+(1-2i)y = 3+4i (iv) (1-i)x+(2+i)y = 4-i (v) (3x-1)+(2y-3)i = 8+7i
Given
(i)
(2x+5)+(y-3)i = 1+2i
2x+5=1 ⇒ x=-2; quad y-3=2 ⇒ y=5
Result
(i)
x=-2, y=5
Given
(ii)
(3x+2)-(4-y)i = 5+3i
3x+2=5 ⇒ x=1; quad -(4-y)=3 ⇒ y=7
Result
(ii)
x=1, y=7
Given
(iii)
(2+i)x+(1-2i)y = 3+4i
(2x+y)+(x-2y)i = 3+4i ⇒ 2x+y=3, x-2y=4
x=4+2y ⇒ 2(4+2y)+y=3 ⇒ 5y=-5 ⇒ y=-1, x=2
Result
(iii)
x=2, y=-1
Given
(iv)
(1-i)x+(2+i)y = 4-i
(x+2y)+(-x+y)i = 4-i ⇒ x+2y=4, -x+y=-1
x=1+y ⇒ (1+y)+2y=4 ⇒ y=1, x=2
Result
(iv)
x=2, y=1
Given
(v)
(3x-1)+(2y-3)i = 8+7i
3x-1=8 ⇒ x=3; quad 2y-3=7 ⇒ y=5
Result
(v)
x=3, y=5