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Assignment 7 MATH 1200 SOLUTION (1) Problem 4 from √ Exercises for Chapter 6 (a) What is i? SOLUTION: Writing i in polar form we have i = cos(2kπ + π2 ) + i sin(2kπ + π2 ) whenever k ∈ Z. So, if α denotes a square root of i then we can write α in polar coordinates as α = r(cos θ + i sin θ). Then α2 = r2 (cos 2θ + i sin 2θ). And if α2 = i then 2θ must be of the form 2kπ + π2 and r2 = 1 so r = 1. Therefore, θ = kπ + π4 for k ∈ Z. This corresponds to two possible √ √ √ √ values for α, namely α = 22 + 22 i (when θ = π4 ) and α = − 22 − 22 i (when θ = π + π4 ) (b) Find all the tenth roots of i. Which one is nearest to i in the Argand diagram? SOLUTION: By the same argument as above: If α10 = i then α must be of the form π cos θ + i sin θ where 10θ = 2kπ + π2 . So θ is of the form 2kπ 10 + 20 for k = 1, 2, 3, .....10 (other values of k correspond to the same complex numbers). 9π 13π 17π 21π 25π 29π 33π 37π 41π I.e, distinct 10th roots of i have value θ = 5π 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 . 9π π And 20 is closest to 2 . √ (c) Find the seven roots of the equation z 7 − 3 + i = 0. Which one of these is closest to the imaginary axis. √ √ 11π SOLUTION: This is equivalent to z 7 = 3−i. And 3−i = 2 cos( 11π 6 + 2kπ) + i sin( 6 + 2kπ) for any k ∈ Z 1 Therefore, as argued above if z satisfies this equation and z = r(cos θ + i sin θ) then r = 2 7 (12k+11)π 35π 47π 59π 71π 83kπ 95π . I.e., θ = 23π and θ must be of the form θ = ( 11π 6 +2kπ)/7 = 42 42 , 42 , 42 , 42 , 42 , 42 , 42 11π (the last of these has the same position as 42 ). The imaginary axis corresponds to the angles π2 and 3π 2 . And the imaginary number cor2π responding to position angle θ = 23π is within radians to the positive imaginary axis 42 42 π 2π ( 23π − = ). And the next closest is the imaginary number corresponding to position 42 2 42 59π 4π 59π 4π angle θ = 42 which is within 42 radians to the negative imaginary axis ( 3π 2 − 42 = 42 ). 23π In any case the root with position angle 42 is closest to the imaginary axis. REMARK: Alternately, on may find the roots in the above problems as described in Example 6.7. I hadn’t yet discussed this example in class, but the general useful result is as follows: Proposition 1. If a + bi ∈ C, then all solutions to the equation z n = a + bi can be found as follows: If α is one solution, and w, w2 , ..., wn are all the nth roots of unity, then z = wα, w2 α, ...., wn α = α are all solutions to the equation z n = a + bi.