1. Problem: Evaluate $$\lim_{z \to \pi i} e^{z^2}$$. Step 1: Substitute $z = \pi i$. Step 2: Compute $z^2 = (\pi i)^2 = \pi^2 i^2 = -\pi^2$ because $i^2 = -1$. Step 3: Evaluate the limit as $e^{-\pi^2}$. 2. Problem: Evaluate $$\lim_{z \to 0} \frac{z^2 + \bar{z}^2}{\operatorname{Re}(z) + \operatorname{Im}(z)}$$ where $z = x + iy$. Step 1: Write $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$ and $\bar{z}^2 = (x - iy)^2 = x^2 - y^2 - 2ixy$. Step 2: Sum numerator: $z^2 + \bar{z}^2 = 2(x^2 - y^2)$, imaginary parts cancel. Step 3: Denominator: $\operatorname{Re}(z) + \operatorname{Im}(z) = x + y$. Step 4: We approach $(x,y) \to (0,0)$. Step 5: Evaluate limit along $y=x$: numerator $2(x^2 - x^2)=0$, denominator $x+x=2x$, ratio $0/0$ indeterminate. Step 6: Evaluate along $y=-x$: numerator $2(x^2 - (-x)^2)=2(x^2 - x^2)=0$, denominator $x - x=0$, ratio $0/0$ indeterminate. Step 7: Try $y=0$: numerator $2(x^2 - 0)=2x^2$, denominator $x + 0= x$, ratio $2x^2/x = 2x \to 0$. Step 8: Try $y=x/2$: numerator $2(x^2 - (x/2)^2) = 2(x^2 - x^2/4) = 2(3x^2/4) = \frac{3}{2}x^2$, denominator $x + x/2 = \frac{3}{2}x$, ratio $\frac{3}{2}x^2 / \frac{3}{2}x = x \to 0$. Step 9: Since along different paths limit is 0, limit is 0. 3. Problem: Evaluate $$\lim_{z \to 0} \frac{e^z - e^{\bar{z}}}{\operatorname{Im}(z)}$$. Step 1: Write $z = x + iy$, $\bar{z} = x - iy$. Step 2: $e^z = e^{x+iy} = e^x (\cos y + i \sin y)$. Step 3: $e^{\bar{z}} = e^{x - iy} = e^x (\cos y - i \sin y)$. Step 4: Compute numerator: $e^z - e^{\bar{z}} = e^x (\cos y + i \sin y) - e^x (\cos y - i \sin y) = e^x (2i \sin y)$. Step 5: Denominator: $\operatorname{Im}(z) = y$. Step 6: Limit becomes $$\lim_{(x,y)\to (0,0)} \frac{e^x (2i \sin y)}{y} = \lim_{(x,y)\to (0,0)} e^x 2i \frac{\sin y}{y}$$. Step 7: $\lim_{x\to 0} e^x = 1$, $\lim_{y\to 0} \frac{\sin y}{y} = 1$. Step 8: Therefore limit is $2i$. 4. Problem: Evaluate $$\lim_{z \to 1 + i} \log_e \sqrt{n^2 + 9^2 + i k e^n} \cdot \frac{n^5}{n}$$ (assuming $n$ is real and $z$ is not directly affecting expression). Step 1: Simplify inside: $\frac{n^5}{n} = n^4$. Step 2: Calculate $\sqrt{n^2 + 81 + i k e^n}$, which depends on $n$ and $k$, no $z$ limit needed. Step 3: Because problem is ambiguous for variable $z$, assume substituting $n=1, k=1$ as typical values. Step 4: Evaluate inside square root: $1 + 81 + i\cdot1 \cdot e^1 = 82 + i e$. Step 5: Compute logarithm and multiply by $n^4=1$. Step 6: Result is $\log_e \sqrt{82 + i e}$. Step 7: Express square root as exponent $1/2$ power: $$\frac{1}{2}\log_e (82 + i e)$$. 5. Problem: Evaluate $$\lim_{z \to 2 - i} (z^2 - z)$$. Step 1: Substitute $z=2 - i$. Step 2: Calculate $z^2 = (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i -1 = 3-4i$. Step 3: Compute $z^2 - z = (3-4i) - (2 - i) = 3 - 4i - 2 + i = 1 - 3i$. 6. Problem: Evaluate $$\lim_{z \to 1 + i} \frac{z^2 -1}{z^2 -1}$$. Step 1: Expression is $\frac{z^2 -1}{z^2 -1}$. Step 2: For $z \neq \pm 1$, fraction is $1$. Step 3: At $z = 1 + i$, $z^2 - 1 \neq 0$; numerator and denominator are equal, so limit is $1$. 7. Problem: Evaluate $$\lim_{z \to e^{i\pi/4}} \left(z + \frac{i}{2}\right)$$. Step 1: Substitute $z = e^{i\pi/4} = \cos (\pi/4) + i \sin (\pi/4) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$. Step 2: Compute sum: $\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} + \frac{i}{2} = \frac{\sqrt{2}}{2} + i \left(\frac{\sqrt{2}}{2} + \frac{1}{2}\right)$. 8. Problem: Evaluate $$\lim_{z \to 1} \frac{z^{4\theta 1}}{z+1}$$ assuming $\theta 1$ is an angle or parameter. Step 1: Substitute $z=1$. Step 2: Compute numerator $1^{4 \theta 1} = 1$. Step 3: Denominator $1 + 1 = 2$. Step 4: Limit is $\frac{1}{2}$. 9. Problem: Evaluate $$\lim_{z \to 2 + i} \frac{z^2 - (2 + i)^2}{z - (2 + i)}$$. Step 1: This is a difference quotient, equivalent to derivative of $z^2$ at $z = 2 + i$. Step 2: Derivative $\frac{d}{dz} z^2 = 2z$. Step 3: Substitute $z = 2 + i$. Step 4: Limit is $2 (2 + i) = 4 + 2i$. 10. Problem: Evaluate $$\lim_{z \to \infty} \frac{z^2 + iz - 2}{(1 + 2i) 2z}$$. Step 1: Divide numerator and denominator by $z$. Step 2: Numerator $\frac{z^2}{z} + \frac{iz}{z} - \frac{2}{z} = z + i - \frac{2}{z}$. Step 3: Denominator $(1 + 2i) 2$ is constant. Step 4: As $z \to \infty$, $-2/z \to 0$, but $z \to \infty$, so numerator grows unbounded. Step 5: Limit is $\infty$, diverges. 11. Problem: Evaluate $$\lim_{z \to \infty} \frac{i z + 1}{2 z - i}$$. Step 1: Divide numerator and denominator by $z$. Step 2: Numerator $i + \frac{1}{z}$, denominator $2 - \frac{i}{z}$. Step 3: As $z \to \infty$, $\frac{1}{z} \to 0$, so limit is $\frac{i}{2}$. Final answers: 1) $e^{-\pi^2}$ 2) $0$ 3) $2i$ 4) $\frac{1}{2} \log_e (82 + i e)$ 5) $1 - 3i$ 6) $1$ 7) $\frac{\sqrt{2}}{2} + i\left( \frac{\sqrt{2}}{2} + \frac{1}{2} \right)$ 8) $\frac{1}{2}$ 9) $4 + 2i$ 10) Diverges to infinity 11) $\frac{i}{2}$