1 - y . i. Since z = x + iy , ( iz + 2) = 2-y . Thus ( iz +...

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Homework 1 From Gamelin : Complex Analysis January 21, 2007 Throughout this assignment we make use of the notation z = x + yi where x, y R , i.e. ± z = x , ² z = y . 1 I.1 1. a. | z - 1 - i | = 1, so | ( x - 1) + i ( y - 1) | = 1. Squaring both sides gives ( x - 1) 2 +( y - 1) 2 =1 , Which is the equation for a circle of radius 1 centered at the point (1 , 1), or 1 + i in complex notation. e. | z - 1 | < | z | , so squaring both sides (which doesn’t change the inequality since both sides are non-negative) gives ( x - 1) 2 + y 2 = | z - 1 | 2 < | z | 2 = x 2 + y 2 . This reduces to - 2 x +1 < 0, or x< - 1 2 . With no restriction on the imaginary part
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Unformatted text preview: y . i. Since z = x + iy , ( iz + 2) = 2-y . Thus ( iz + 2) &gt; 0, means 2 &gt; y , so we are left with the part of the plane below the line y = 2. 3. Letting a = u + iv , and z = x + iy as usual, we have | z | 2-2 ( az ) + | a | 2 = x 2 + y 2-2 (( u-iv )( x + iy )) + u 2 + v 2 = x 2 + y 2-2 ( ux + vy + i ( uy-vx )) + u 2 + v 2 = x 2 + y 2-2( ux + vy ) + u 2 + v 2 = ( x-u ) 2 + ( y-v ) 2 . 1...
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This note was uploaded on 07/27/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.

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