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Compute \((1 + i)^4\)

 Aug 16, 2019
 #1
avatar+140 
+2

Binomial theorem expansion so we have \(\sum _{i=0}^4\binom{4}{i}\cdot \:1^{\left(4-i\right)}i^i=1+4i-6-4i+1=\boxed{-4}~~\blacksquare\)

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 Aug 16, 2019
 #2
avatar+140 
+2

Note that i is additionally the index of the summation.

 Aug 16, 2019
 #3
avatar+23273 
+2

Compute  \(\large{\left(1 + i\right)^4}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\left(1 + i\right)^4} \\ &=& \Big(~\left(1 + i\right)^2~\Big)^2 \quad &| \quad \left(1 + i\right)^2 = 1+2i+i^2 \\ &=& \Big(~1+2i+i^2~\Big)^2 \quad &| \quad i^2=-1 \\ &=& \left(~1+2i-1~\right)^2 \\ &=& \left( 2i \right)^2 \\ &=&4i^2\quad &| \quad i^2=-1 \\ &=&4(-1) \\ &=& \mathbf{-4} \\ \hline \end{array}\)

 

laugh

 Aug 16, 2019
 #4
avatar+6045 
+1

\(\text{Just as an alternative way to calculate this}\\ z = 1+i = \sqrt{2} e^{i \pi/4}\\ z^4 = (\sqrt{2})^4 e^{4(i\pi/4)} = 4e^{i\pi} = -4\)

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 Aug 16, 2019

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