# Order of operation when dividing

One of the first things I learned at math at secondury school is the order of operations:

1. Things between brackets
2. Multiplication and division

(At that time exponents and roots wheren't introduced.)

Now at university, I've seen formulas like $$Z_C = 1/j\omega C$$ (for the impedance of a capacitor in an electric circuit) which is also written as $$Z_C = \frac{1}{j\omega C}$$

But since both multiplication and division has the same rank on the order of operation, I would interpreted (if I wouldn't know better) the first formula as $$Z_C = 1/j\omega C = \frac{1}{j} \omega C = \frac{\omega C}{j}$$ (which could also be written as $-j\omega C$ since $j$ is the imaginary unit, but that's beside the point).

This is just one example, another would be $$R(\lambda, T) = \frac{2\pi h c^2}{\lambda^5 (e^{hc/\lambda kT} - 1)}$$ for Plank blackbody law, where it is meant that $hc/\lambda kT = \frac{hc}{\lambda k t}$

My question is, does division has higher priority than multiplication in the order of operation? I know my question sounds really basic, but I'm never told the answer.

• If the first is intended to be $\frac{1}{j\omega C}$ then it is very sloppy notation. Unfortunately, your best bet is to figure it out from context because it should as you say be interpreted correctly as the second, not the first. That being said, if it was intended to be the second, you would think they would have just written $\omega C/j$ instead... the fact that they didn't should be a clue that they are sloppy and might not have written correctly. – JMoravitz Nov 3 '16 at 19:08
• Just use common sense!! – john Nov 4 '16 at 9:10

Thus, in $\frac{1}{jωC}$ the multiplicative operations of $j*w*C$ will be prioritized. Factoring out $(\frac1j)wC$ will simply lead to an incorrect answer.
For any number represented in form $\frac xy$, where $x$ and $y$ are placeholders for any number of different mathematical operations:
you must complete all operations on the numerator $x$ and the denominator $y$ FIRST.