FA07Ma1aSol9

# FA07Ma1aSol9 - MATH 1A FALL 2007 HOMEWORK 9 SOLUTIONS...

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MATH 1A FALL 2007 HOMEWORK 9 SOLUTIONS Problem 1. [Apostol, page 365, problems 3,e,h,i,j ] Compute the modulus and principal argument of each of the following complex numbers e) - 3 + 3 i h) ( - 1 - i ) 3 i) 1 / (1 + i ) j) 1 / (1 + i ) 2 Solution: Note that the modulus and principal argument of x + iy are r = x 2 + y 2 and θ = arctan ( y/x ) ( - π, π ], also the n-th power of z = re is z n = r n e inθ . Use the above formulas, we immediately get the following: e) r = 2 3 , θ = 5 π/ 6 f) r = 2 2 , θ = - π/ 4 g) r = 2 / 2 , θ = - π/ 4 h) r = 1 / 2 , θ = - π/ 2 Problem 2. [Apostol, page 365, problems 5,a,b,d,f ]Make a sketch showing the set of all z in the complex plane which satisfy each of the following equations a) | z | < 1 b) z + ¯ z = 1 d) | z - 1 | = | z + 1 | f) z + ¯ z = | z | 2 Solution: Let z = x + iy a) | z | < 1 means that x 2 + y 2 < 1, that is the unit disc without boundary. b.) z + ¯ z = 1 means that 2 x = 1, thus is the vertical line Re ( z ) = 1 / 2. d.) | z - 1 | = | z + 1 | means that the distance between z and -1 is equal to the distance between z and 1, i.e. z is on the line which goes through the midpoint

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