Step 1: For (I), substitute the line \(y=mx\) directly into the original expression instead of simplifying algebraically first. With \(y=mx\), \[f(x,mx)=\frac{x^2-m^2x^2+2imx^2}{x^2+m^2x^2}=\frac{x^2(1-m^2+2im)}{x^2(1+m^2)}=\frac{1-m^2+2im}{1+m^2},\] which is independent of \(x\) and depends only on the slope \(m\).
Step 2: For \(m=0\) (the x-axis) this gives \(1\); for \(m=1\) (the line \(y=x\)) it gives \((0+2i)/2=i\). Since approaching along different straight lines through the origin gives different values, the limit as \(z\to0\) does not exist. This confirms (I) is false using a family of straight-line paths rather than the polar substitution.
Step 3: For (II), apply the sequential criterion for limits: if \(\lim_{z\to0}f(z)\) existed, every sequence \(z_n\to0\) would give the same limit \(f(z_n)\).
Step 4: Take \(z_n=1/n\to0\); then \(f(z_n)=\operatorname{Re}(1/n)/|1/n|=1\) for every \(n\), so this sequence suggests limit \(1\). Now take \(w_n=i/n\to0\); then \(f(w_n)=\operatorname{Re}(i/n)/|i/n|=0/(1/n)=0\) for every \(n\), suggesting limit \(0\). Since the two sequences give different results (\(1\) versus \(0\)), the limit does not exist, so \(f\) is not continuous at \(z=0\) even though \(f(0)=1\) was assigned.
Step 5: Both statements fail. \[\boxed{\text{Neither (I) nor (II) is true}}\]
Match List-I with List-II and choose the correct option:
| LIST-I (Function) | LIST-II (Value) |
|---|---|
| (A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: |z-a|=r, r > 0 \) | (III) \( 2i\pi \) |
| (B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \) | (IV) \( i\pi \) |
| (C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: |z|=3 \) | (II) \( 2i\pi(e^4 - e^2) \) |
| (D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: |z|=2 \) | (I) \( -4 + 2i\pi \) |
Choose the correct answer from the options given below: