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May 8th, 2014, 04:30 AM   #1
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How to Solve this?

On Simplification, [i^19 + (1/25)^25]^2

becomes

(a) -4
(b) 4-1
(c) 2+i
(d) None Of the above
sagarerande is offline  
 
May 8th, 2014, 05:34 AM   #2
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d
let x= i^19, y = (1/25)^25
[x + y]^2
x^2 + 2xy + y^2
The y^2 bit is going to be a very small number
The 2xy bit is a bunch of i
The x^2 bit is -1
Add those and it can't be a,b or c
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