I was solving a problem, and I found this:
$$\frac{3}{i}(2 - i)=\frac{6}{i} - 3$$
According to symbolab, I must multiply by the conjugate of the denominator, then the result is:
$$=\frac{6\cdot (-i)}{i \cdot (-i)} -3= -6i - 3$$
Should a complex number always be multiplied by its conjugate? Because if I had multiplied only by the positive $i$, the result would have been $6i - 3$, which is different.
On
If you had multiplied both numerator and denominator by $i$, it would give you the same result in this case:
$$ \frac{6}{i} - 3 = \frac{6\cdot i}{i \cdot i} - 3 = \frac{6i}{-1} - 3 = -6i - 3 $$
But when you try to rationalize a complex denominator that has both a real and imaginary part, then you have to multiply by its conjugate in order to make the denominator real, e.g.:
$$ \frac{6}{3 - 2i} - 3 = \frac{6(3+2i)}{(3-2i)(3+2i)} - 3 = \frac{18+12i}{13} - 3 = \frac{18 + 12i - 3\cdot 13}{13} = \frac{-21 + 12i}{3}$$
The result would NOT have been different, it would have still been $-6i-3$.
If you are multiplying by $+i$, then:
$\dfrac 3i(2-i)=\dfrac{3(2-i)}i=\dfrac{6-3i}i=\dfrac{6-3i}i\cdot\dfrac ii=\dfrac{i(6-3i)}{-1}=-(6i-3i^2)=-(6i+3)=-6i-3$
Note: If there is addition or subtraction in the denominator, meaning if complex numbers in the form $a+bi$ where $a,b\ne0$ are in the denominator, you have to multiply by the conjugate.