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Problem of the week #22 - August 27th, 2012

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Jameson

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Jan 26, 2012
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Thank you to soroban for proposing this problem!

\(\displaystyle \left| (1+2i)^n \right|^2\) for n=1,2,3... can be generalized in a very simple form that doesn't include any notation related to complex numbers.

1) Find a way to generalize the nth term.
2) Prove your generalization is valid

Hint 1:
Start with n=1. Squaring the magnitude of a complex number cancels out the square root normally in that calculation so given $z = a+bi$ we are looking at the square of the magnitude, or $a^2+b^2$. The magnitude is \(\displaystyle \sqrt{a^2+b^2}\) but take note of the outer exponent, 2.

For $n \ge 2$ you'll have to simplify $(1+2i)^n$ to the form of $a+bi$ and then calculate $a^2+b^2$.


Hint 2:
Remember that for complex numbers, z and w, |zw|=|z||w|. You can apply this to the problem through \(\displaystyle (1+2i)^n=(1+2i)^{n-1}(1+2i)\)


If this problem seems too tricky for you then I suggest reading up on complex numbers, focusing on multiplying them and finding the magnitude.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) Reckoner
2) Sudharaka

Solution (from Reckoner):
Calculating \(\left|(1+2i)^n\right|^2\) for a few small values of \(n\) suggests a simple pattern:

For \(n=1\),
\[\left|(1+2i)^n\right|^2 = |1+2i|^2 = 5.\]

For \(n=2\),
\[\left|(1+2i)^n\right|^2 = \left|(1+2i)^2\right|^2 = \left|1+4i+4i^2\right|^2\] \[ = |-3+4i|^2 = 25 = 5^2.\]

For \(n=3\),
\[\left|(1+2i)^n\right|^2 = \left|(1+2i)^3\right|^2 = \left|1+6i+12i^2+8i^3\right|^2\] \[=|-11 - 2i|^2 = 125 = 5^3.\]

This suggests that \(\left|(1+2i)^n\right|^2 = 5^n\). For proof, we use induction. Note that the base case \(n=1\) was already shown above. Assume that, for some positive integer \(k\), \(\left|(1+2i)^k\right|^2 = 5^k\). Then, setting \(n = k+1\) gives

\[\left|(1+2i)^{k+1}\right|^2 = \left|(1+2i)^k(1+2i)\right|^2\]

\[= \left|(1+2i)^k\right|^2\cdot|1+2i|^2,\]

which, by the inductive hypothesis,

\[=5^k\cdot|1+2i|^2 = 5^k\cdot5 = 5^{k+1}\]

as required. Hence, \(\left|(1+2i)^n\right|^2 = 5^n\) for all positive integers \(n\).
 
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