Answer to Question #5125 in Calculus for moha

Question

(1+ i )^32 ... Give the answer in the form z = a + bi.

Answer 1

Let's use binomial theorem:

(1+i)^32 = (32,0)*1^32*i^0 + (32,1)*1^31*i^1 + ... + (32,32)*1^0*i^32 =
= | 1^k=1 | =
= (32,0)*i^0 + (32,1)*i^1 + ... + (32,32)*i^32 =
= | i^0 = 1; i^k=-1 if k is even number, and i^k=i if k is odd number | =
= (32,0) + (32,1)*i - (32,2) + (32,3)*i - (32,4) + ... - (32,32) =
= | grouping | =
= [(32,0) - (32,2) - (32,4) - ... - (32,32)] + [(32,1) - (32,3) - ... - (32,31)]i =
= | (32,k)=(32,32-k) | =
= [ 2( (32,0) + (32,2) + ... +(32,14) ) + (32,16) ] + [(32,1) - 2( (32,3) + (32,5) + ... + (32,15) )]i =
= | calsulations | =
= [ 2(1 + 496 + 35960 + 906192 + 10518300 + 64512240 + 225792840 + 471435600) + 601080390] +
& + [ 32 - 2(4960 + 201376 + 3365856 + 28048800 + 129024480 + 347373600 + 565722720)]i =
= 2147483648 - 2147483584i.