Question

Find
limz→iz2−iz−i

Answer 1

Answer on Question #58319 – Math – Complex Analysis

Question

Find $\lim_{z \to i} z^2 - i z^{-i}$

Solution

As $z^2 - i z^{-i} = e^{2 \ln z} - i e^{-i \ln z}$

and in the vicinity of $i$ we can write down $\ln z = \ln r + i\theta$.

Thus,

$$\lim_{z \to i} z^2 - i z^{-i} = \lim_{\theta \to \frac{\pi}{2}} \lim_{r \to 1} \left( e^{2(\ln r + i\theta)} - i e^{-i(\ln r + i\theta)} \right)$$

Finally we get

$$\lim_{\theta \to \frac{\pi}{2}} \left( e^{2i\theta} - i e^{\theta} \right) = e^{i\pi} - i e^{\frac{\pi}{2}} = -1 - i e^{\frac{\pi}{2}} = -1 - i \sqrt{e^{\pi}}$$

Answer: $\lim_{z \to i} z^2 - i z^{-i} = -1 - i \sqrt{e^{\pi}}$.

Question

Find $\lim_{z \to i} z^2 - i z - i$

Solution

$$\lim_{z \to i} z^2 - i z - i = \lim_{z \to i} z^2 - \lim_{z \to i} i z - \lim_{z \to i} i = i^2 - i \cdot i - i = -i$$

Answer: $\lim_{z \to i} z^2 - i z - i = -i$.

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