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

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#### Jameson

Staff member
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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#### Jameson

Staff member
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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