Let z = x + iy
∴ Re(z) = x and Im(z) = y
Given that f(z) = |Re z Im Z|1/2
∴ f(z) = |xy|1/2 = √x√y + i0.
By comparing f(z) = u + iv, we get
u(x, y) = √x√y and v(x, y) = 0
∴ u(0, 0) = √0 x √0 = 0
Now, ux(0, 0) = \(\underset{h\rightarrow0}{Lim}\) \(\frac{u(h,0)-u(0,0)}{h}\)
= \(\underset{h\rightarrow0}{Lim}\) \(\frac{\sqrt h\times0-0}{h}\) = 0
uy(0, 0) = \(\underset{k\rightarrow0}{Lim}\) \(\frac{u(0,k)-u(0,0)}{k}\)
= \(\underset{k\rightarrow0}{Lim}\) \(\frac{0\times\sqrt k-0}{k}\) = 0
Also, since v(x, y) = 0
∴ vx = 0 and vy = 0
Therefore, ux = vy = 0
And uy = -vx = 0
Hence, function f(z) satisfies cauchy reimann equation.
Now, since, f(z) = √x√y and z = x+iy = (x, y).
∴ f(x, y) = √x√y
fx(0, 0) = \(\underset{h\rightarrow0}{Lim}\) \(\frac{f(h,0)-f(0,0)}{h}\) = \(\underset{h\rightarrow0}{Lim}\) \(\frac{\sqrt h\times0-0}{h}\) = 0
fy(0, 0) = \(\underset{k\rightarrow0}{Lim}\) = \(\frac{f(0,k)-f(0,0)}{k}\) = \(\underset{k\rightarrow0}{Lim}\) \(\frac{0\times\sqrt k-0}{k}\) = 0
∴ f'(0, 0) = \(\underset{(h,k)\rightarrow(0,0)}{Lim}\) \(\frac{f(0+h,0+k)-f(0,0)-h\,f_x(0,0)-k\,f_y(0,0)}{\sqrt{h^2+k^2}}\)
= \(\underset{(h,k)\rightarrow(0,0)}{Lim}\) \(\frac{\sqrt h\sqrt k-0}{\sqrt{h^2+k^2}}\)
= \(\underset{h\rightarrow0}{Lim}\) \(\frac{\sqrt h\sqrt{mh}}{\sqrt{h^2+m^2h^2}}\) (By taking k = mh)
= \(\underset{h\rightarrow0}{Lim}\) \(\frac{h\sqrt{m}}{h\sqrt{1+m^2}}\)
= \(\frac{\sqrt{m}}{\sqrt{1+m^2}}\) depends on m.
∴ f(x, y) is not differentiable at origin (0, 0).
Thus, f(z) satisfies cauchy reimann equation but not differentiable at origin (0, 0).