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# HELP ASAP HELP!!

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129
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Simplify (1+i)^2016-(1-i)^2016

Oct 6, 2019

#1
+19879
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I believe the answer is  0 .

Oct 6, 2019
#2
0

How did you get that?

I don't understand how to do it.

Guest Oct 6, 2019
#3
+2547
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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.

CalculatorUser  Oct 7, 2019
edited by CalculatorUser  Oct 7, 2019
#4
+107116
+1

Simplify (1+i)^2016-(1-i)^2016

$$1+i=\sqrt2e^{\frac{\pi}{4}i} \qquad \text{When presented in polar form}\\~\\ (1+i)^{2016}=\left( \sqrt2e^{\frac{\pi}{4}i} \right)^{2016}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{\pi i} )^{504}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{2\pi i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(e^{0 i} )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*(1 )^{252}\\ (1+i)^{2016}= (\sqrt2)^{2016}*1\\~\\ (1-i)^{2016}=\left( \sqrt2e^{\frac{-\pi}{4}i} \right)^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{\frac{-\pi}{4}i})^{2016}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{-2\pi i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} (e^{0 i})^{252}\\ (1-i)^{2016}=(\sqrt2)^{2016} *1\\ so\\ (1+i)^{2016}-(1-i)^{2016}=(\sqrt2)^{2016}-(\sqrt2)^{2016}=0$$

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Oct 7, 2019