Complex no. sum (1 Viewer)

[juantheron]

 

[qrpw]

I think I have made a mistake somewhere, but is the answer just r^3 where r=modulus.

 

[math man]

yes the answer is
the easiest way to do it is as follows, let

since mods are all equal. Now

since a + bc is real, Im(a +bc)=0, therefore:

now working abc is as follows:

 

[math man]

just a side note to the above, for these types of questions it beneficial to work with mod arg form as it has a lot
nicer properties than cartesian form, such as just adding args when multiplying. The algebra involved in say letting a = x+iy would be very tedious.
So for future questions you would want to use mod arg form over cartesian any day

 

[jenslekman]

I think I have made a mistake somewhere, but is the answer just r^3 where r=modulus.

keep up the good work

 

[qrpw]

Thanks math man, that was kinda what I did (and I found my mistake). But wouldn't
instead of

 

[lolcakes52]

This is the same problem i ran into.

 

[math man]

Yes you are right, I proved it but it was
a long process in doing so. Basically I used
fact that the three numbers there were real
to deduce a relation in terms of theta alpha
and beta and then subbed it into abc.
After a lot of trig manipulation I showed
That the a cancels out I the abc and the
I equate that equation with bc and proved
the same result for theta

 

[juantheron]

Thanks friends but answer given is