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granice ciagow licz zespolonych

: 8 maja 2008, o 01:02
autor: profesorq
(a) \(\displaystyle{ z_n=(\frac{1+i}{2})^n}\)
(b)\(\displaystyle{ z_n=(\sqrt{n^2+\sqrt{n}}-n)e^{in}}\)
(c)\(\displaystyle{ z_n=\frac{3ni+i^n}{n-i}}\)

granice ciagow licz zespolonych

: 9 maja 2008, o 23:58
autor: Lorek
a)
\(\displaystyle{ |z_n|=\left|\left(\frac{1+i}{2}\right)^n\right|=\left(\frac{\sqrt{2}}{2}\right)^n \\\lim_{n\to\infty}|z_n|=0\Rightarrow \lim_{n\to\infty}z_n=0}\)