Question

Show the function f defined by f(z)=|Re z Imz|^1/2 satisfy the cauchy riemann equation at the origin ,but is not differentiable at this point.

Answer 1

Let z = x + iy

∴ Re(z) = x and I_m(z) = y

Given that f(z) = |Re z I_m 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, u_x(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

u_y(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

∴ v_x = 0 and v_y = 0

Therefore, u_x = v_y = 0

And u_y = -v_x = 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

f_x(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

f_y(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).