2010/027) What is the square root of -i

THis can be done both with and with out De Moivre's formula as mentioned in

http://en.wikipedia.org/wiki/De_Moivre%2…

1st with it

-i = 0 + (-1) i = r cos t + r sin t

r^2 = 1

cos t = 0 and sin t = - 1 so t = 3pi/2

so -i = e^(3pi/2) i

so sqrt(i) = e^(3pi/4) i or e^5pi/4 taking (3pi/2 + 2npi/2) n = 0 and 1 n =2 3 gives 1st value

= cos 3pi/4 + i sin 3 pi/4 or cos 5pi/4 + i sin 5 pi/4

= - 1/sqrt(2) + 1/sqrt(2) i or 1/sqrt(2) - 1/sqrt(2) i

in case you are not familiar with De Moivre's formula then say

sqrt(-i) = (a+ib)

sqare both sides

-i = a ^2-b^2 + 2abi

so a^2-b^2 -= 0
and 2ab = -1
a^2-b^2 = 0 => a = +/-b so ab = -1/2 and we get a = 1/sqrt(2) b= - 1/sqrt(2)

or a = -1/sqrt(2) b= + 1/sqrt(2)

giving same results

2010/026) Strong induction problem

Suppose x is a real number, x does NOT equal 0, and (x + 1/x) is an integer

Prove for all n ≥ 1, x^n + 1/(x^n) is an integer.
x+ 1/x is in integer = n (given)

so (x+ 1/x)^2 = n^2

x^2 + 1/x^2 = n^2-2 is an integer
so true for 1 => true for 2

let it be true for upto k
now x^k+ 1/x^k

(x^k+ 1/x^k)(x+ 1/x)

= x^(k+1) + 1/x^(k-1) + x^(k-1) + 1/x^(k+1)

so x^(k+1) + 1/x^(k+1) = (x^k+1/x^k)(x+1/x) - (x^(k-1) + 1/x^(k-1))

if it is true for all n upto k so RHS is integer then LHS is integer so for k+ 1 so the induction step is proved

hence proved