How do I find the complex conjugate of #12/(5i)#?

2 Answers
Aug 12, 2015

Answer:

The conjugate is #(12i)/5#

Explanation:

To find a conjugate of a complex number we first have to convert it to a form #a+bi#. To do this here we can multiply both numerator and denominator by #i#

#z=12/(5i)=(12i)/(5i^2)=(12i)/(-5)=-(12i)/5#

Now to calculate the conjugate we just have to change sign of the imaginary part:

#barz=(12i)/5#

Aug 13, 2015

In #a + bi# form, this starts off as:

#0 + 12/(5i)#

Note that #[12/(5i) = 12/5*1/i] != [(12i)/5 = 12/5 i]#)

#12/(5i) * (i/i) = (12i)/(5i^2) = (12i)/(-5)#

We currently have:

#a + bi = 0 + (12i)/(-5)#

The conjugate is #a - bi#, thus we get:

#a - bi = 0 - (12i)/(-5)#

#= (12i)/(5) = color(blue)(12/5 i)#