Question

Where is the function f(z) =|z|^2+I (x-iy) +1 differentiable at?

Answer 1

Where is the function f(z) =|z|^2+I (x-iy) +1 differentiable at?

If z=x+i.y, where x and y are real, we are given that f(z) = (x^2 +y^2) + i(x-i.y) + 1 = {x^2 +y^2+y +1} + i.x = u(x,y) +i.v(x,y) say, where u and v are real valued functions of two real variables.

 

The partial derivatives u_x = 2x, u_y = 2y+1, v_x = 1, and v_y = 0, for every x and y. Hence we see that the first Cauchy-Riemann condition u_x = v_y holds only when x=0, and second condition u_y=-(v_x) holds only when 2y+1= -1 ==> y=(-1).

 

Therefore as the functiond u and v have continuous partial derivatives everywhere, the function f(z) is differentiable only at the point 0+i.(-1) = (-i) and the value

 

f(-i) = 0^2+(-1)^2+(-1)+1 +i.0 = 1 and the value f'(-i) of the derivative there is

 

(u_x)(0,-1) + i(v_x)(0,-1) = i.

i