DotKite said:Homework Statement
find the limit z_n = [(1+i)/sqrt(3)]^n as n -> ∞.
Homework Equations
3. The Attempt at a Solution
Apparently the limit is zero (via back of the book), but I have no clue how they got that answer.
(1 + i)^n seems to be unbounded, thus i do not see how z_n can go to zero I am lost.
Dick said:(1+i)^n is unbounded. But its absolute value is |1+i|^n. What's that??
DotKite said:|1+i|^n = [sqrt(2)]^n?
Dick said:Sure, so can you show the limit of |z_n| is 0? That would show the limit of z_n is also 0.
But it is true if the limit is zero. ##|z_n| \rightarrow 0## if and only if ##z_n \rightarrow 0##.DotKite said:that is not generally true,
take |(-1)^n + i/n| which converges to 1
{(-1)^n + i/n} does not converge.
A limit of a complex sequence is the theoretical value that a sequence of complex numbers approaches as the number of terms in the sequence increases. In simpler terms, it is the value that the terms in the sequence get closer and closer to, but may never actually reach.
The limit of a complex sequence can be calculated by finding the limit of both the real and imaginary parts of the sequence separately. If both limits exist, then the limit of the complex sequence is the complex number formed by combining the real and imaginary limits.
Studying the limit of a complex sequence allows us to understand the behavior and properties of sequences of complex numbers. It also has practical applications in areas such as signal processing, electrical engineering, and physics.
Some common methods for finding the limit of a complex sequence include using the definition of a limit, the squeeze theorem, and the ratio test. Other methods such as L'Hôpital's rule and the Cauchy product can also be used in certain cases.
One common misconception is that the limit of a complex sequence must be a complex number. In reality, the limit may also be infinity or not exist at all. Another misconception is that the limit of a complex sequence is always equal to the last term in the sequence, when in fact this is not always the case.