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How can I prove that $$\lim _{z \to i} \frac {iz^3-1}{z+i} =0$$ Using the epsilon-delta definition I get stuck after factorizing $iz^3-1$ to get $$\left|\frac {(1+iz)(z^2+iz-1)}{z+i}\right| < \epsilon $$ Usually what I would do is get a $|z-i|$ term somewhere and then try to bound the other terms (at least that is what I did with real limits) how can I go around this problem? What am I doing wrong?

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  • $\begingroup$ A general method one can use (for this problem there is simpler ways): The numerator (which is a qubic polynomial) vanish for $z=i$ so we can write it on the form $iz^3 - 1 = a(z - i)^3 + b(z-i)^2 + c(z-i)$ for some constant $a,b,c$. Since the leading term is $iz^3$ we must have $a=i$. The rest of the constants can be figured out by expanding and comparing to the left hand side (or using calculus; comparing derivatives of the two sides at $z=i$). $\endgroup$ – Winther May 12 '16 at 22:22
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Factoring $i$ out of $1+zi$ gives $i(z+\dfrac{1}{i})=i(z-i)$.

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From @user247327's hint, which is to factor the term $i$ out of $1+zi$, results in $$\left|1+iz\right| = \left|i(z+\dfrac{1}{i})\right| = \left|i\right|\left|z-i\right| = \left|z-i\right|.$$

Therefore, $$ \left|\frac {(1+iz)(z^2+iz-1)}{z+i}\right| = \left|z-i\right|\left|\dfrac{z^2+iz-1}{z+i}\right|. $$

From this, try to bound the numerator and denominator of $\left|\dfrac{z^2+iz-1}{z+i}\right|$ by finding an appropriate disc of radius $\delta$, i.e. determine $\delta>0$ such that $$ 0<\left|z-i\right|<\delta={}\operatorname{min}\left\{1,\text{some expression of }\epsilon\right\} $$ and $\left|\dfrac{(1+iz)(z^2+iz-1)}{z+i}\right| < \epsilon$.

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