- #1
complexnumber
- 62
- 0
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
f(z) = z^2 = (x + iy)^2 = x^2 + 2ixy - y^2
u(x,y) = x^2 - y^2
v(x,y) = 2xy
The image of x+y=1 under f(z) = z^2 refers to the set of all complex numbers that result from plugging in x+y=1 into the function f(z) = z^2.
To find the image of x+y=1 under f(z) = z^2, you can simply plug in x+y=1 into the function and simplify the resulting expression.
The image of x+y=1 under f(z) = z^2 is significant because it represents the output or result of the function for a specific input. It helps us visualize how the function transforms or maps the input to the output.
Yes, the image of x+y=1 under f(z) = z^2 is unique. This means that for any given input, there can only be one output or image.
Yes, the image of x+y=1 under f(z) = z^2 can be a complex number, as the function f(z) = z^2 operates on complex numbers. This means that the resulting image can also be a complex number.