Mod and Arg?? (1 Viewer)

[asianese]

We had a question in our assessment on Thursday:

0 <= |arg(z-1+root3)| <= pi/3

I didn't see the modulus signs and proceeded with the argument.

I have not seen these types before but one of the teachers said their class had done it...wtf

 

[cutemouse]

We had a question in our assessment on Thursday:

0 <= |arg(z-1+root3)| <= pi/3

I didn't see the modulus signs and proceeded with the argument.

I have not seen these types before but one of the teachers said their class had done it...wtf

I think that's the same as |arg(z-1+root3)| <= pi/3 because by definition |arg(z-1+root3)|>=0 for all z (abs value of anything is non negative).

So the expression becomes -pi/3 <= arg(z-1+root3) <= pi/3.

 

[asianese]

Ergh I just hope everyone gets it wrong then ==