As others explained, you can't use DC math to solve the problem: you have to consider

the impedances as vector quantities containing both reactance and resistance.

But the length of the stub (assuming it gives an exact match) is sufficient to solve

the problem. You can follow along and change any of the parameters as needed

if I have made a wrong assumption.

Let's assume that you are using RG-213 coax for the stub. Using VK1OD's

Transmission

Line Loss Calculator for 18' of RG-213 terminated by 0.01 ohms (close to a short circuit)

at 1.8 MHz, the impedance looking into the stub is 0.55+j16.21 ohms, or .002098-j0.0616 S

in parallel form. You should get basically the same result using a coil with a reactance of 16.21

ohms (about 1.3uH) in place of the stub. (Any analysis that doesn't also work using such a

coil in place of the coax stub is suspect. The residual 0.55 ohms of resistance is due to the

imperfect short circuit and the cable losses.)

Now if this gives you a perfect match to 0.02 S (50 ohms) it must be that the impedance

at the end of the coax is (0.02 +j0 S) - (0.002098 - j0.0616 S) = 0.017992 + j0.0616 S,

or 4.37 - j14.96 ohms, so the SWR without the stub should be about 12.5 : 1. (If you don't

have a calculator that goes between parallel and series notation, you can plug one or the

other into VK1OD's calculator with a very short bit of transmission line.)

In fact, knowing the impedance at the transmitter, VK1OD's calculator will give you the

actual impedance at the antenna if you know the type and length of feedline.

Why did the analyzer give an impedance of 350 ohms for this load? I have no idea. Perhaps

it is out of the range over which the analyzer is accurate, or the stub didn't really give a

perfect match to 50 ohms. (There is no guarantee that it will for random impedances.)

So all that we did was to put 0.55+j16.21 ohms in parallel with 4.37 - j14.96 ohms to get

50 ohms. We know the voltage across the stub from the output power (since it is at

a 50 ohm point in the system). If we used a coil instead, the losses would be based on

the Q of the coil. The losses will be somewhat higher using a coax stub due to the

losses in the coax.

But power is ONLY dissipated in the resistive portion of impedance, not in the reactive

part. That's why you can't use the total impedance to calculate power lost.

(And say "Hi" to your club members from me.)