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Find the value of (1+i)^16

 Oct 7, 2018
 #1
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Simplify the following:
(i + 1)^16

Compute (1 + i)^16 by repeated squaring. For example a^7 = a a^6 = a (a^3)^2 = a (a a^2)^2.
(1 + i)^16 = ((i + 1)^8)^2 = (((i + 1)^4)^2)^2 = ((((i + 1)^2)^2)^2)^2:
((((i + 1)^2)^2)^2)^2

Expand (i + 1)^2.
(i + 1)^2 = 1 + i + i - 1 = 2 i:
(((2 i)^2)^2)^2

Evaluate (2 i)^2.
(2 i)^2 = -4:
((-4)^2)^2

Evaluate (-4)^2.
(-4)^2 = 16:
16^2

Evaluate 16^2.
=256

 Oct 7, 2018
 #2
avatar+689 
+2

To write what he just said in a way that is easier to see,

 

\((1+i)^{16} = ((i+1)^{8})^{2} = (((i+1)^{4})^{2})^{2} = ((((i+1)^{2})^{2})^{2})^{2}\)

 

Now if we expand and solve for \((i+1)^{2}\), we get:

 

\(i^2+2i+1 = -1+2i+1 = 2i\)

 

Therefore, \(((((i+1)^{2})^{2})^{2})^{2} = (((2i)^2)^2)^2\)

 

Since \(i = -\sqrt1\)  ,   \((2i)^2 = -4\)

 

Therefore, \((((2i)^2)^2)^2 = ((-4)^2)^2\)

 

Evaluating \(((-4)^2)^2\), we get \((16)^2\)

 

\(16^2=256\), so \(\boxed{256}\) is our answer

 

smiley

 Oct 7, 2018

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