+2862 I'm not sure I haven't taken algebra 2 yet, but I think its like htis
Since 2016 is even,
what is (1+i)^2?
What is (1-i)^2
So I think th answer should be 4i, after plugging that in the calc. But EP may be correct he probably took alg 2 already.
+118609 Simplify (1+i)^2016-(1-i)^2016
\(1+i=\sqrt2e^{\frac{\pi}{4}i} \qquad \text{When presented in polar form}\\~\\ (1+i)^{2016}=\left( \sqrt2e^{\frac{\pi}{4}i} \right)^{2016}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{\pi i} )^{504}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{2\pi i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{0 i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(1 )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*1\\~\\ (1-i)^{2016}=\left( \sqrt2e^{\frac{-\pi}{4}i} \right)^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{\frac{-\pi}{4}i})^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{-2\pi i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{0 i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} *1\\ so\\ (1+i)^{2016}-(1-i)^{2016}=(\sqrt2)^{2016}-(\sqrt2)^{2016}=0 \)
