Why must I use the conjugate of $3i$ for $\frac{5i-4}{3i}$ to get rid of $i$ in the denominator instead of just using $i$? My algebra textbook says to multiply the numerator and denominator by the conjugate of the denominator to eliminate $i$. For $$\frac{5i-4}{3i}$$
The conjugate of the denominator is $-3i$ and after doing all the calculations the final results is $$\frac{5}{3}+\frac{4}{3}i$$
If the example had a complex number in the denominator containing both the real portion and imaginary portion, then of course multiply by the conjugate. 
But for this particular case where the real portion of the complex number in the denominator is $0$, why would it be considered "incorrect" or "improper" to multiply the numerator and denominator by $i$, and more "proper/correct" to always use the conjugate instead? 
Multiplying by the numerator and denominaotor by $i$ doesn't change the results in this case. 
 A: When you are told to multiply the numerator and denominator by the complex conjugate of the denominator, it is because you are being given a method that works in all cases.
If $c+di$ is the denominator and $c,d$ are real, then multiplying the numerator and denominator by $-c+di$ will also work. And in $\dfrac{3+7i}{2+6i},$ multiplying the numerator and denominator by $1-3i$ also works.
Multiplying both the numeraotor and the denominator by the conjugate of the denominator is sufficient for the problem, but not necessary in all cases.
(I've seen instances where a student saw something like $\dfrac{6+8i} 2$ and multiplied by the numerator and the denominator by the complex conjugate of $2.$ That is silly.)
A: If the real part is zero in the denominator, you do not need multiply by the conjugate. you just write
$$\frac{5i-4}{3i}=\frac{5i}{3i}-\frac{4}{3i}=$$
$$\frac{5}{3}+\frac{4i}{3}$$
using the fact that
$$\frac{1}{i}=-i$$
A: This works for any denominator with real part zero, since $$\frac{x+iy}{wi}=\frac1i\left(\frac xw+\frac ywi\right)=\frac yw-\frac xwi$$ the imaginary number can be taken out via its reciprocal.
When we have a denominator with a real part other than zero, it becomes impossible to simply take out the imaginary part as shown below: $$\frac{x+iy}{a+bi}=\frac1i\left(\frac x{\frac ai+b}+\frac y{\frac ai+b}i\right)$$ leaves us with a non-real denominator still. This is because we can't cater for the real when we deal singly with the imaginary, and vice versa.
Hence we would need to use the conjugate, since $z\overline z=|z|^2$ is real.
A: Or as an alternative
$$\frac{5i-4}{3i}=\frac{5i-4}{3i}\frac i i=\frac{-5-4i}{-3}=\frac53+\frac43 i$$
A: If you consider a different example:
$$\frac{-4+5i}{1+3i}$$
the only way to do the same thing is multiplying by the complex conjugate:
$$\frac{-4+5i}{1+3i}\cdot\frac{1-3i}{1-3i}$$
A: You can multiply by (and only multiply by) any nonzero real multiple of your complex conjugate. If $a+bi \neq 0$ is your denominator you can multiply by $c+di$ where $ad+bc=0$ and $ac-bd \neq 0$.

First suppose $a \neq 0$ . Then:
$$-\frac{b}{a}c=d$$
So,
$$(c,d)=(c,\frac{-b}{a}c)=\frac{c}{a} (a,-b)$$
Hence we see that $c+di$ must be a nonzero real multiple of the complex conjugate $a-bi$. We can check this leads to $ac-bd \neq 0$ as long as $a+bi$ is not zero. 

Next, suppose $a=0$ but $a+bi \neq 0$. Then we need $bc=0$. Since $b \neq 0$ we have $c=0$. We may choose any value of $d$. So,
$$(c,d)=(0,d)=-\frac{d}{b}(0,-b)$$
Again we see that we need $c+di$ to be a nonzero real multiple of the complex conjugate $a-bi$. 
A: You can factor out the real part of the denominator for consistency:
$$\frac{5i-4}{3i}=\frac13\cdot \left(\frac{5i-4}{i}\right)=\frac13\cdot \left(\frac{5i-4}{i}\cdot \frac{-i}{-i}\right)=\frac13\cdot \left(\frac{5+4i}{1}\right)=\frac53+\frac43i.$$
Also, it is similar to freeing the denominator from irrationality. Compare $\frac{5\sqrt{3}-4}{3\sqrt{3}}$ versus $\frac{5\sqrt{3}-4}{3\sqrt{3}-1}$.
So, there is a general rule and a particular rule. Sometimes textbooks state both, sometimes not. Simple example: the general root formula of quadratic equation $ax^2+bx+c=0$ is:
$$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$$
and when $b$ is even, the particular root formula is:
$$x_{1,2}=\frac{-\left(\frac b2\right)\pm \sqrt{\left(\frac b2\right)^2-ac}}{a}.$$
