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Unformatted text preview: Complex Analysis Spring 2001 Homework I Solution 1. Conway, Chapter 1, section 3, problem 3. Describe the set of points satisfying the equation  z a    z + a  = 2 c , where c > and a R. To begin, we see from the triangle inequality that 2  a  =   2 a  =  ( z a ) ( z + a )   z a    z + a  = 2 c and so the set is empty in case  a  < c. The case where  a  = c corresponds to equality in the triangle inequality, so z a and z + a are nonnegative multiples of each other. This occurs when z  a  or z  a  . However, only one of these rays satisfies the original equation, since the distance from z to a must be greater than the distance from z to a. From this we see that the ray starting at a directed away from a is the solution set. We may now assume  a  > c. Rewriting the equation as  z a  = 2 c +  z + a  , squaring both sides, and collecting terms gives the equation 4 ax = 4 c 2 + 4 c  z + a  . Solve for  z + a  and square again to get ( x + a ) 2 + y 2 = c 2 + 2 ax + a 2 c 2 x 2 . After gathering terms, this becomes x 2 c 2 a 2 c 2 y 2 = c 2 which is the equation of a hyperbola. Only one of the two branches satisfies the original equation though, since again z must be farther from a than from a. Thus the branch with focus at a is the solution set. If a is permitted to be complex, then the set of points is again a ray or a branch of a hyperbola. This may be seen by rotating a and z counterclockwise by arg ( a ). The absolute value is invariant under rotation, so the form of the equation remains the same for the new variable. In the new variable, a lies on the real axis, so the previous analysis applies. Rotatation transforms a ray to a ray, and the branch of a hyperbola to a branch of a hyperbola. 2. The equation az + b z + c = 0 can have solution set consisting of a single point, or a line in the complex plane. (a) Find necessary and sufficient conditions on the complex constants a, b , and c so that the solution set is a line. First, if either a = 0 or b = 0, the system has either a unique solution (when one of the two is equal to zero but not the other), no solutions (when a = b = 0 and c 6 = 0), or all z C in the case that all coefficients are zero. Thus we suppose that neither a nor b is 0. If we separate the real and imaginary parts of the equation, writing a = a 1 + ia 2 , b + b 1 + ib 2 ,and c = c 1 + ic 2 then we obtain two real equations ( a 1 + b 1 ) x + ( b 2 a 2 ) y = c 1 ( a 2 + b 2 ) x + ( a 1 b 1 ) y = c 2 ....
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This note was uploaded on 10/22/2009 for the course MATH 552 taught by Professor Snider during the Spring '01 term at USC.
 Spring '01
 SNIDER

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