Impedance problem (polar to rectangular)

Impedance matching is particularly important if you need maximum power transfer, and the applications for this are many, ranging from audio outputs to antenna drive.

In a high speed and RF world, impedance matching is necessary to prevent signal reflection; this becomes an issue if the physical distance between a driver and receiver is greater than about 1/10 of a signal rise or fall time (rule of thumb: the strict limit is far smaller). This is part of transmission line theory and application.

Another RF application is matching an antenna (which is normally in free space with an impedance of 377 ohms) to the internal transceiver circuitry (which is commonly at 50 or 75 ohms), as maximum power transfer is particularly important here, for both transmit and receive operation.

Providing none of these apply (such as a low frequency amplifier chain), then generally a receiver input impedance should be much larger than the source impedance - once more, a rule of thumb is a factor of 10.

The above assumes voltage amplifiers. Current amplifiers (a smaller but very important class of amplifiers) have low input impedance and high output impedance and are worthy of a discussion in their own right.

HTH

According to my notes the total impedance is 1002 Ω, -86.6°

Your notes are correct and I think the mistake you may have made is adding 60 ohm (resistive) to the 1000 ohms (reactive) to get 1060 ohms. You need to add them as squares then take the square root hence \$\sqrt{1000^2 + 60^2}\$ = 1001.8 ohms.

The 1000 ohms is the reactive impedance of C in parallel with L. I'm presuming you know how to derive this because you appear to be making the right noises in your question i.e. the "(Xc*Xl)/(Xc+Xl)" bit is correct.