Huaizhong R. answered 06/04/25
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The function is neither injective nor surjective.
- f takes the same value at x+iy and x–iy, so it is not injective.
- the imaginary part of f is y2–1, which means f only takes value in the half-plane above the line z=–1.
- To determine the range of f, let f(x+iy)=u+iv, where u, v are real. Then 3x+2=u, and y2–1=v. We have x=(u-2)/3, and y=±√(v+1). We can see easily that u can take any real value but v has to be ≥ -1. So the range of f is the half-plane above the line v=-1(inclusive).