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How do I prove the following limit using the limit definition? $$\lim_{z\to i} z^2=-1$$ Using the limit definition $$|z^2+1|<\epsilon, \;\text{whenever} 0<|z-i|<\delta$$ so I factor out to get that $$|z^2+1|=|z+i||z-i|<\epsilon$$ which means that if I am able to bound $$|z+i|<K \tag{1}$$ then I can say that $$|z-i||z+i|<K|z-i|<\epsilon \tag{2}$$ and from there I could proceed to proving the limit, the only problem is that I do not know how to go from $(1)$ to $(2)$ using complex numbers. How could I prove this limit?

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  • $\begingroup$ Your missing step is $|z+i|<K$, which is easy to achieve when $z$ is bounded. $\endgroup$ – Yves Daoust Apr 28 '16 at 6:53
  • $\begingroup$ To prove (1): use the triangle inequality on $|z+i | = |z-i+2i|$, or simplier, by using Yvea argument... To go from (1) to (2) just use that $|ab|=|a||b|$ even for complex numbers (both usual real/imaginary part splitting and polar coordinates work) $\endgroup$ – b00n heT Apr 28 '16 at 7:00
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By taking $0<\delta<\min\left\{1,\frac{\epsilon}3\right\}$ we get \begin{align*} |z^2+1|&=|z+i||z-i|\\ &\le\left(|z-i|+|2i|\right)|z-i|\qquad\text{from the triangular inequality}\\ &<(\delta +2)\delta\\ &<3\cdot\frac{\epsilon}3 \end{align*} provided $|z-i|<\delta$.

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Let $\delta = \min\{1,\frac{\epsilon}{3}\}$, then $$ 0<|z-i|<1 $$ and so $$ |z|-1<1 $$ by triangle inequality. Thus, $$ |z+i|\le |z|+1< 3 $$ by triangle inequality again.

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Consider a neighborhood of radius $R$ around $i.$ Then $$|z^2+2|=|z+i||z-i|\le (|z-i|+|2i|)|z-i)\le (R+2)|z-i|.$$

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