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Question

Evaluate (1+i2)8+(1i2)8

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Solution

Consider the given equation,

(1+i2)8+(1i2)8=28

L.H.S.,

(1+i2)8+(1i3)8=[(1+i2)2]4+[(1i3)2]4

=[(12+i222+2.1.i2)2]4+[(12+i2322.1.i3)2]4

i=12

=[(1+(1)2+2..i2)]4+[(1+(1)32i3)]4

=[(1+2.i2)]4+[(22i3)]4

=[(1+2.i2)2]2+[(22i2)2]2

=((1)2+22.i22242)2+((2)2+(2i3)2+8i3)2

i2=1

=(1842)2+(412+8i3)2=(74i2)2+(8+8i3)2

=49+(32i2)+28i2+64+384i21283

=113+416.i2+282.i1283.i

=113+416.i2+282.i1283.i

i2=1

=113+416(1)+282.i1283.i

=306+282.i1283.i


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