1. ## complex numbers

find the indicated power of the complex number (1+ i) to the 5th power

2. Originally Posted by scrappy
find the indicated power of the complex number (1+ i) to the 5th power
$(1+i)^5$

This can be done in a number of ways. The most direct way would probably be to expand the 1+i using the binomial expansion:
$(1+i)^5 = \left ( \begin{array}{c} 5 \\ 0 \end{array} \right )1^5i^0 + \left ( \begin{array}{c} 5 \\ 1 \end{array} \right )1^4i^1 + \left ( \begin{array}{c} 5 \\ 2 \end{array} \right )1^3i^2$ $+ \left ( \begin{array}{c} 5 \\ 3 \end{array} \right )1^2i^3 + \left ( \begin{array}{c} 5 \\ 4 \end{array} \right )1^1i^4 + \left ( \begin{array}{c} 5 \\ 5 \end{array} \right )1^0i^5$

$(1+i)^5 = 1 + 5i + 10i^2 + 10i^3 + 5i^4 + i^5$

$(1+i)^5 = 1 + 5i - 10 - 10i + 5 + i$

$(1+i)^5 = -4 - 4i$

-Dan

3. Hello, scrappy!

. . Multiply it out . . .

Find: . $(1+ i)^5$

$(1 + i)^5\;=\;1^5 + 5(1^4)(i) + 10(1^3)(i^2) + 10(1^2)(i^3) + 5(1)(i^4) + i^5$

. . . . . . $= \;1 + 5(1)(i) + 10(1)(-1) + 10(1)(-i) + 5(1)(1) + i$

. . . . . . $= \;1 + 5i - 10 - 10i + 5 + i$

. . . . . . $= \;-4 - 4i$

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If you are familiar with DeMoivre's Theorem, it's faster.

Since $1 + i \:=\:\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

. . we have: . $\left[\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\right]^5\;=\;\left(\sqrt{2}\right)^5\left[\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right]$

Then: . $4\sqrt{2}\left[\text{-}\frac{1}{\sqrt{2}} + i\left(\text{-}\frac{1}{\sqrt{2}}\right) \right] \;=\;-4 - 4i$