Image of x+y=1 under f(z) = z^2

Thread starter complexnumber
Start date Jul 31, 2009
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To find the image of the line x + y = 1 under the transformation f(z) = z^2, the line can be expressed in the complex plane as z = x + i(1 - x). Substituting this into f(z) gives f(z) = (x + i(1 - x))^2, resulting in u(x,y) = x^2 - (1 - x)^2 and v(x,y) = 2x(1 - x). This leads to the equations for the real and imaginary parts of the transformed points. The discussion focuses on how the quadratic transformation affects the line in the complex plane, ultimately revealing the nature of the image created by f(z).

Jul 31, 2009

complexnumber

For f(z) = z^2 find the image of x + y = 1

f(z) = z^2 = (x + iy)^2 = x^2 + 2ixy - y^2

u(x,y) = x^2 - y^2
v(x,y) = 2xy

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Jul 31, 2009

Mark44

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Here's how I would approach this problem. Every point on the line has coordinates (in the complex plane) of (x, 1 - x). A point on this line can be represented as z = x + i(1 - x). What does your function f do to these complex numbers?

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