Լուծեք {l}{{(1+i/1-i)}^m=1}{j=2}{k=j}{l=k}{n=l}{o=n}{p=o}{q=p}{r=q}{s=r}{text{Solvefor}ttext{where}}{t=s}

\left. \begin{array} { l } { {(\frac{1 + i}{1 - i})} ^ {m} = 1 }\\ { j = 2 }\\ { k = j }\\ { l = k }\\ { n = l }\\ { o = n }\\ { p = o }\\ { q = p }\\ { r = q }\\ { s = r }\\ { \text{Solve for } t \text{ where} } \\ { t = s } \end{array} \right.

Լուծել m, j, k, l, n, o, p, q, r, s, t-ի համար

t=2

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Complex Number

5 խնդիրները, որոնք նման են.

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You may find the following easier to understand. Let |a| represents the order of a and let the g.c.d. of |a| and k be g. (a^k)^\frac{|a|}{g}=(a^{|a|})^\frac{k}{g}=e and so |a^k| ...

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(I'll do it for dimension n because the difficulty is the same) Suppose you write \{e_1,\ldots,e_n\} for the canonical basis (could be any basis, actually), and let \{f_1,\ldots,f_n\} be another ...

If a_n = \frac{{(n + 1)}^n}{n^{n - 1}} and c > 0 then \lim_{n \to \infty} n^{1-c}(a^c_n-a^c_{n-1}) =ce^c

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I think the approach in my answer to the question which you have linked should work without much change. To simplify typing I will use A = a_{n}, B = a_{n - 1}. Note that A/n, B/n \to e. I first ...

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