# Complex numbers, help

Watch
Announcements
#1

This is my first step using Euler’s formula

Not sure what do do next to get it into the standard modulus argument form of r(cos(a) isin(a)) which seems to be the right way to go (worked for Part A).
Would appreciate a little help! Perhaps there is an identity I could apply or something else?
Last edited by Maximus 190; 1 month ago
0
1 month ago
#2
so to find -z you can think of it as a rotation of 180 degrees around the origin ?
0
1 month ago
#3
for part B consider the top of the fraction as 1 + 0i

when you divide complex numbers you divide the moduli ( here 1 / 4 ), and subtract the arguments.... here 0 - α
0
#4
(Original post by the bear)
so to find -z you can think of it as a rotation of 180 degrees around the origin ?
Okay. So surely shouldn’t 1/z give the same argument so it’s theta but that is incorrect?
0
#5
(Original post by the bear)
for part B consider the top of the fraction as 1 + 0i

when you divide complex numbers you divide the moduli ( here 1 / 4 ), and subtract the arguments.... here 0 - α
Oh okay that is cool! Thanks
1
#6
(Original post by the bear)
so to find -z you can think of it as a rotation of 180 degrees around the origin ?
What about the angle for Part C?

The complex conjugate is just a reflection in the real axis so i*sin(a) becomes -i*sin(a) which equals i*sin(a pi) but it says that’s wrong?
Last edited by Maximus 190; 1 month ago
0
1 month ago
#7
(Original post by Maximus 190)
What about the angle for Part C?

The complex conjugate is just a reflection in the real axis so i*sin(a) becomes -i*sin(a) which equals i*sin(a pi) but it says that’s wrong?
The best way to view it is, as you say, a reflection in the x-axis. You've done a rotation by 180 degrees which is different. You want
a -> -a
Then map it to the range you want (positive angle, measured anticlockwise from the positive x-axis).

The sin bit is correct but you've forgotten to reason about the real part of z which is cos().
Last edited by mqb2766; 1 month ago
0
1 month ago
#8
(Original post by Maximus 190)

Not sure what do do next to get it into the standard modulus argument form of r(cos(a) isin(a)) which seems to be the right way to go (worked for Part A).
Would appreciate a little help! Perhaps there is an identity I could apply or something else?

Last edited by RDKGames; 1 month ago
1
#9
(Original post by mqb2766)
The best way to view it is, as you say, a reflection in the x-axis. You've done a rotation by 180 degrees which is different. You want
a -> -a
Then map it to the range you want (positive angle, measured anticlockwise from the positive x-axis).

The sin bit is correct but you've forgotten to reason about the real part of z which is cos().
Oh yeah I forgot changing the sin angle will change the cos and hence the real part too :P got it now thanks

(Original post by RDKGames)

Oh yeah you’re right that is a bit simpler, thanks
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (105)
60.34%
No (69)
39.66%