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# Write z1 x1 iy1 and z2 x2 iy2 then z1 z22 x1x22

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Unformatted text preview: Proof We square both sides and compare them. Write z1 = x1 + iy1 and z2 = x2 + iy2. Then |z1 + z2|2 = (x1+x2)2 + (y1+y2)2 = x12 + x22 + 2x1x2 + y12 + y22 + 2y1y2. On the other hand, (|z1| + |z2|)2= |z1|2 + 2|z1||z2| + |z2|2 = x12 + x22 + y12 + y22 + 2|z1||z2|. Subtracting, (|z1| + |z2|)2 - |z1 + z2|2 = 2|z1||z2| - 2(x1x2 + y1y2) = 2[|(x1,y1)||(x2,y2)| - (x1,y1).(x2,y2)] (in vector form) = 2[|(x1,y1)||(x2,y2)| - |(x1,y1)||(x2,y2)| cos å] = 2 |(x1,y1)||(x2,y2)| (1 - cos å ) ≥ 0, giving the result. Note The triangle inequality can also be seen by drawing a picture of z1 + z2. 5. We now consider the polar form of these things: If z = x+iy, we can write x = r cosø and y = r sinø, getting z = r cosø + ir sinø, so This is called the polar form of z . It is important to draw pictures in order to feel comfortable with the polar representation. Here r is the magnitude of z, r = |z|, and ø is called the argument of z, denoted arg(z). To calculate ø, we can use the fact that tan ø = y/x. Thus ø is not arctan(y/x) as claimed in the book, but by:...
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## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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