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X r ⊆ Y 1 , . . , Y s , (1) if each of X1 , . . , Xr is a linear combination of Y1 , . . , Ys . 1) Suppose that each of X1 , . . , Xr is a linear combination of Y1 , . . , Ys and that each of Y1 , . . , Ys is a linear combination of X1 , . . , Xr . Then by (a) equation (1) above X1 , . . , X r ⊆ Y 1 , . . , Y s and Y1 , . . , Y s ⊆ X 1 , . . , X r . Hence X1 , . . , X r = Y 1 , . . , Y s . 1) Suppose that each of Z1 , . . , Zt is a linear combination of X1 , . . , Xr . Then each of X1 , .

Also Arg (4 + i) = α, where tan α = 1/4 and 0 < α < π/2. Hence α = tan −1 (1/4). 1 = 2 (−3)2 + (−1)2 y ✻ ✛ 1√ 9+1= = 2 √ 10 . 2 Also Arg (−1 + 2i) = π − α, where tan α = 2 and 0 < α < π/2. Hence α = tan −1 2. 60 ✲x α ✏ ✏✏ ✮ −3−i 2 ❄ Also Arg ( −3−i ) = −π + α, where 2 1 3 tan α = 2 / 2 = 1/3 and 0 < α < π/2. Hence α = tan −1 (1/3). (iii) The number lies in the second quadrant of the complex plane. √ | − 1 + 2i| = (−1)2 + 22 = 5. 4+i ✲x ❄ (ii) The number lies in the third quadrant of the complex plane.

12 12 (d) z5 w2 = 25 (cos ( 5π − 2π ) + i sin ( 5π 32 4 6 4 32 11π 11π (cos 12 + i sin 12 ). 9 = − 2π )) 6 (a) (1 + i)2 = 2i, so (1 + i)12 = (2i)6 = 26 i6 = 64(i2 )3 = 64(−1)3 = −64. √ )2 = −i, so (b) ( 1−i 2 1−i √ 2 −6 = 1−i √ 2 = (−i)−3 = 2 −3 −1 1 −1 = = = −i. 3 i −i i √ √ 8. (i) To solve the equation z 2 = 1 + 3i, we write 1 + 3i in modulus– argument form: √ π π 1 + 3i = 2(cos + i sin ). 3 3 Then the solutions are π π √ + 2kπ + 2kπ 3 zk = 2 cos + i sin 3 , k = 0, 1. 2 2 62 Now k = 0 gives the solution z0 = √ √ √ π π 2(cos + i sin ) = 2 6 6 3 i + 2 2 √ 3 i =√ +√ .

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