I actually did read this thread before posting haha, but converting to Cartesian seems so unnecessary, there must be a better method? What about if the moduli of both numbers was the same:

re^{i\theta}

and

re^{i\phi}

Reply 3

the bear

Original post by lizard54142

Say we had two complex numbers:

re^{i\theta}

and

se^{i\phi}

, how would we find the midpoint of these two numbers?

I think I am correct in saying that the argument would be

\dfrac{\theta + \phi}{2}

, but is there a straightforward to find the modulus also?

i think this only works if the moduli are equal ?

Reply 4

lizard54142

Original post by the bear

i think this only works if the moduli are equal ?

True, my bad.

Reply 5

ghostwalker

Original post by lizard54142

What about if the moduli of both numbers was the same:

re^{i\theta}

and

re^{i\phi}

A minor bit of geometry will give you this special case - have a go.

Reply 6

rayquaza17

Original post by lizard54142

I actually did read this thread before posting haha, but converting to Cartesian seems so unnecessary, there must be a better method? What about if the moduli of both numbers was the same:

re^{i\theta}

and

re^{i\phi}

But we define the argument to be between -pi and pi so maybe this won't work?

Sorry I've got my differential equations head on ATM for my exam tomorrow so I might be wrong.

Posted from TSR Mobile

Reply 7

lizard54142

Original post by rayquaza17

But we define the argument to be between -pi and pi so maybe this won't work?

Sorry I've got my differential equations head on ATM for my exam tomorrow so I might be wrong.

Posted from TSR Mobile

I'm almost certain if they both have the same modulus, the argument is just the average of both their arguments... we can then just adjust this to fit the required range.

Reply 8

lizard54142

Original post by ghostwalker

A minor bit of geometry will give you this special case - have a go.

The argument would just be the average of their arguments... not sure about the modulus :s-smilie:

Reply 9

the bear

Original post by lizard54142

The argument would just be the average of their arguments... not sure about the modulus :s-smilie:

you could use the sine rule probably

Reply 10

ghostwalker

Original post by lizard54142

The argument would just be the average of their arguments... not sure about the modulus :s-smilie:

Draw a diagram, and do a little geometry - I know you can do it.

Reply 11

lizard54142

Original post by ghostwalker

Draw a diagram, and do a little geometry - I know you can do it.

Okay, I got off my a*se and did a diagram, and came up with:

w=rcos(\frac{\theta - \phi}{2})

, where w is the modulus of the midpoint and

\theta > \phi

Reply 12

ghostwalker

Original post by lizard54142

Okay, I got off my a*se and did a diagram, and came up with:

w=rcos(\frac{\theta - \phi}{2})

, where w is the modulus of the midpoint and

\theta > \phi

In general, yes. You will need a slight modification to deal with some angles to avoid getting a negative value for the modulus. Modulus signs will do.

And as rayquaza17 pointed out, you may have an issue with the argument as well.

But you can check the cases.