How do you find the conjugate of -2/i^3?

**viccal** · August 15th 2012, 02:01 AM

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

**a tutor** · August 15th 2012, 02:14 AM

$\frac{-2}{i^3}=\frac{-2i}{i^4}=-2i$ .

**HallsofIvy** · August 15th 2012, 05:39 AM

To find the conjugate of any complex number, replace i with -i. You could "rationalize the denominator" first, as "a tutor" did:
$\frac{-2}{i^3}= -2i$ and then its conjugate is $-2(-i)= 2i$ . Or you can immediately write that the conjugate is $\frac{-2}{(-i)^3}= \frac{2}{i^3}$ . If you were then to multiply both numerator and denominator by i you would again get $\frac{2i}{i^4}= \frac{2i}{(i^2)^2}= \frac{2i}{(-1)^2}= 2i$ .

**Deveno** · August 15th 2012, 05:12 PM

aw...i was gonna say find its minimal real polynomial, and grab the other root :P

**viccal** · August 18th 2012, 06:11 PM

What's the reason you multiplied the numerator and denominator by i?

**skeeter** · August 18th 2012, 07:09 PM

Originally Posted by viccal

What's the reason you multiplied the numerator and denominator by i?

tutor used the fact that $i^4 = 1$ in order to get $\frac{-2}{i^3}$ in the form $a \pm bi$ in an efficient manner.

he ended up with $-2i = 0 - 2i$ ... note that the conjugate is $0 + 2i = 2i$

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