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Question Number 177071 by cortano1 last updated on 30/Sep/22

Commented by Frix last updated on 30/Sep/22

$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by Frix last updated on 30/Sep/22

$$\underset{z\to -\mathrm{i}}{\lim }\frac{z^2+3\mathrm{i}z-2}{z+\mathrm{i}}=\underset{z\to -\mathrm{i}}{\lim }\frac{\cancel{(z+\mathrm{i})}(z+2\mathrm{i})}{\cancel{z+\mathrm{i}}}=\mathrm{i}$$

Answered by Frix last updated on 30/Sep/22

$$z=r{\mathrm{e}}^{\mathrm{i}\theta }-\mathrm{i}$$$$\frac{z^3+3\mathrm{i}z-2}{z+\mathrm{i}}=$$$$=(r^2\cos 2\theta +3r\sin \theta -3+\frac{\cos \theta +\sin \theta }{r})+$$$$+(r^2\sin 2\theta -3r\cos \theta +3+\frac{\cos \theta -\sin \theta }{r})\mathrm{i}$$$$z\to -\mathrm{i}\iff r\to 0\mathrm{and}\mathrm{the}\mathrm{red}\mathrm{terms}\mathrm{are}\mathrm{undefined}$$$$\Rightarrow \mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

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