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Unformatted text preview: Mat104 Solutions to Problems on Complex Numbers from Old Exams (1) Solve z 5 = 6 i . Let z = r (cos + i sin ). Then z 5 = r 5 (cos 5 + i sin 5 ). This has modulus r 5 and argument 5 . We want this to match the complex number 6 i which has modulus 6 and infinitely many possible arguments, although all are of the form / 2 , / 2 2 , / 2 4 , / 2 6 , / 2 8 , / 2 10 , . . . . (We will see that we dont really lose anything if we drop the in our list of possible arguments for 6 i .) So we choose r 5 = 6 and 5 = 2 or 2 + 2 or 2 + 4 or 2 + 6 or 2 + 8 or 2 + 10 or . . . In other words, in order to have z 5 = 6 i we should take z of the form r (cos + i sin ) where r = 5 6 and = 10 or 10 + 2 5 or 10 + 4 5 or 10 + 6 5 or 10 + 8 5 or 10 + 10 5 or . . . Notice that there are only really 5 choices for theta. The sixth choice = / 10 + 10 / 5 = / 10+2 gives the same complex number as the first choice, where we simply take = / 10. So there are exactly 5 solutions to z 5 = 6 i corresponding to r = 5 6 and = / 10 , / 10 + 2 / 5 , / 10 + 4 / 5 , / 10 + 6 / 5 and / 10 + 8 / 5. If we sketch these complex numbers we would see that they all lie on the circle of radius 5 6 1 . 43 and they are separated from each other by an angle of 2 / 5. (2) Find the real part of (cos 0 . 7 + i sin 0 . 7) 53 . This is the same as ( e . 7 i ) 53 = e 53 . 7 i = e 37 . 1 i = cos(37 . 1) + i sin(37 . 1) . So the real part is simply cos(37 . 1). (3) Find all complex numbers z in rectangular form such that ( z 1) 4 = 1....
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 Fall '07
 Nelson
 Complex Numbers

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