Understanding Complex Geometric Sequences: A Revision Question

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SUMMARY

The discussion focuses on understanding complex geometric sequences, specifically the sequence defined by \( u_n = 3(1+i)^{n-1} \). It establishes that the product sequence \( v_n = u_n u_{n+k} \) also forms a geometric sequence, with the ratio \( \frac{v_n}{v_{n-1}} = (1+i)^2 \). The derivation of these sequences illustrates the properties of complex numbers in geometric progression, confirming that both \( u_n \) and \( v_n \) maintain consistent ratios across their terms.

PREREQUISITES
  • Understanding of complex numbers, specifically \( i \) as the imaginary unit.
  • Familiarity with geometric sequences and their properties.
  • Knowledge of the formula for the sum of geometric sequences.
  • Basic algebraic manipulation involving exponents and ratios.
NEXT STEPS
  • Study the properties of complex numbers in sequences and series.
  • Learn about the derivation and application of the sum formula for geometric sequences.
  • Explore advanced topics in complex analysis, focusing on sequences and series.
  • Investigate the implications of geometric sequences in real-world applications, such as finance and population growth.
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Students and educators in mathematics, particularly those focusing on complex analysis and geometric sequences, as well as anyone preparing for exams involving these concepts.

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need a hand with a revision question, I don't quite understand how to go about solving it question is attached below
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Hi, and welcome to the forum!

According to the definition of $u_n$,
\[
u_4=(1+i)u_3=(1+i)^2u_2=(1+i)^3u_1=3(1+i)^3=-6+6i.
\]
In general, $u_n=3(1+i)^{n-1}$. From there
\[
v_n=u_nu_{n+k}=3(1+i)^{n-1}\cdot3(1+i)^{n+k-1}=9(1+i)^{2n+k-2}.
\]
In particular,
\[
v_{n+1}=9(1+i)^{2n+k}=(1+i)^2v_n
\]
which means that $v_n$ is also a geometric sequence.

Another way to show this is to note that $\dfrac{u_n}{u_{n-1}}=1+i$ for all $i$. Therefore
\[
\dfrac{v_n}{v_{n-1}}=\dfrac{u_nu_{n+k}}{u_{n-1}u_{n+k-1}}=
\dfrac{u_n}{u_{n-1}}\cdot\dfrac{u_{n+k}}{u_{n+k-1}}=(1+i)(1+i)=(1+i)^2.
\]
The ratio does not depend on $n$; therefore, $v_n$ is a geometric sequence.

To answer other questions, you need to know formulas for the sum of geometric sequences and other information, which you can find in Wikipedia.

Please also read the http://mathhelpboards.com/rules/, especially rules 8 and 11.
 

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