What is the conjugate of $\displaystyle \frac{\ -2}{i^3}$?

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- Aug 15th 2012, 02:01 AMviccalHow do you find the conjugate of -2/i^3?
What is the conjugate of $\displaystyle \frac{\ -2}{i^3}$?

- Aug 15th 2012, 02:14 AMa tutorRe: How do you find the conjugate of -2/i^3?
$\displaystyle \frac{-2}{i^3}=\frac{-2i}{i^4}=-2i$.

- Aug 15th 2012, 05:39 AMHallsofIvyRe: How do you find the conjugate of -2/i^3?
To find the

**conjugate**of any complex number, replace i with -i. You could "rationalize the denominator" first, as "a tutor" did:

$\displaystyle \frac{-2}{i^3}= -2i$ and then its conjugate is $\displaystyle -2(-i)= 2i$. Or you can immediately write that the conjugate is $\displaystyle \frac{-2}{(-i)^3}= \frac{2}{i^3}$. If you were then to multiply both numerator and denominator by i you would again get $\displaystyle \frac{2i}{i^4}= \frac{2i}{(i^2)^2}= \frac{2i}{(-1)^2}= 2i$. - Aug 15th 2012, 05:12 PMDevenoRe: How do you find the conjugate of -2/i^3?
aw...i was gonna say find its minimal real polynomial, and grab the other root :P

- Aug 18th 2012, 06:11 PMviccalRe: How do you find the conjugate of -2/i^3?
What's the reason you multiplied the numerator and denominator by i?

- Aug 18th 2012, 07:09 PMskeeterRe: How do you find the conjugate of -2/i^3?