It is known that for the function 1 2 2 ( ) tan ln y f x iy i x y x + = + there is no way to define f z( ) at any z on the imaginary axis such that f is continuous there

4. It is known that for the function f(x+iy)=tan1(xy)ilnx2+y2 there is no way to define f(z) at any z on the imaginary axis such that f is continuous there. (a) Verify, particularly, that limzif(z) does not exist. [Hint: Choose two simple paths approaching i that give different limits] (b) Using the Cauchy-Riemann equations show that f is analytic on the region where Re(z)=0, then derive that f(z)=zi (consequently, f has an antiderivative either on the domain Re(z)>0 or on the domain Re(z)0 from 1 to 1i. 4. It is known that for the function f(x+iy)=tan1(xy)ilnx2+y2 there is no way to define f(z) at any z on the imaginary axis such that f is continuous there. (a) Verify, particularly, that limzif(z) does not exist. [Hint: Choose two simple paths approaching i that give different limits] (b) Using the Cauchy-Riemann equations show that f is analytic on the region where Re(z)=0, then derive that f(z)=zi (consequently, f has an antiderivative either on the domain Re(z)>0 or on the domain Re(z)0 from 1 to 1i