How to Solve this?

[sagarerande]

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

becomes

(a) -4
(b) 4-1
(c) 2+i
(d) None Of the above

 

[Quaid]

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

becomes

(a) -4
(b) 4-1
(c) 2+i
(d) None Of the above

Hi sagarenrande:

You need to begin to simplify the given expression. If you make a start, you may realize the answer, with some thought. Note that all of the coefficients (in multiple-choice answers a through c) are integers. Your first few steps of simplification will show you whether that form is possible.

Have you studied algebra? Specifically, have you practiced using powers of i before? What is i^2? What is i^3? What is i^4? Keep going, up to i^8.

There is a repeating pattern; do you see it? The pattern gives a way to determine the value of i^19 easily.

And, if you're familiar with the FOIL algorithm for squaring binomials, a little thought about what happens to that 1/25, as it first gets raised to the 25th power (yikes, that's a giant denominator) and then gets squared after that, ought to lead you to the correct conclusion.

So, do some simplification, and see what you think. If you still need help, please show your work, or tell us where you're stuck.

Cheers

 

[HallsofIvy]

To start with, do you know what \(\displaystyle i^{19}\) is? That should be very easy- you only need to observe that 19= 4(4)+ 3.

 

[Dale10101]

For us slower folks

To start with, do you know what \(\displaystyle i^{19}\) is? That should be very easy- you only need to observe that 19= 4(4)+ 3.

i¹⁹ = i⁴⁽⁴⁾⁺³

i⁴ = (i)(i)(i)(i) = (i²)(i²) = (-1)(-1) = 1

i³ = (i)(i)(i) = (i²)(i) = (-1)(i) = -i

i² = (i)(i) = (i²) = -1

i¹ = i

this all you need to know, mod 4, which took me a moment to see but shall now never forget.