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I'd like to prove $\lim_{z\to -2} \frac{5z}{z+i} = 4+2i$. So I was able to get: \begin{align*} \left| \frac{5z}{z+i} - (4+2i)\right| = \left|\frac{(1-2i)z + (2-4i)}{z+i}\right| = \left|\frac{(1-2i)(z+2)}{z+i}\right| \end{align*}

The $|z+2|$ looks familiar since we can bound it with $\delta$ but I am stuck with the rest.

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Given $\epsilon>0$, let $\delta<\min\{\sqrt{5}^{-1}\epsilon,1\}$, for $0<|z+2|<\delta$, then $|z+i|=|z+2+(i-2)|\geq|i-2|-|z+2|=\sqrt{5}-1>2-1=1$, so $\dfrac{|1-2i|}{|z+i|}=\dfrac{\sqrt{5}}{|z+i|}<\sqrt{5}$, and hence $\dfrac{|1-2i||z+2|}{|z+i|}<\sqrt{5}\delta<\epsilon$ for all such $z$.

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In general: given a complex function $f(z)$, $f(z)\rightarrow w$ if and only if $\Re(f(z))\rightarrow \Re(w)$ and $\Im(f(z))\rightarrow \Im(w)$ (i.e. if the real part of the function converges to the real part of the limit and likewise for the imaginary parts). The proof of that is not hard, and once you know that result, you only have to prove convergence of two real sequences, which is more standard.

In this case: you can prove the limit directly. Rewrite: $$\frac{5z}{z+i}=\frac{5z(z-i)}{(z+i)(z-i)}=\frac{5z(z-i)}{z^2+1},$$ and now proving the limit should be straight forward.

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