• Best of luck to the class of 2025 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here

Complex Number Question. HELP.. (1 Viewer)

D

DkAssain

Guest
let w = (3+4i)/5 and z = (5+12i)/13 so that |z|=|w|=1

(i) find wz and w(conjugate z) in the form x+iy.

(ii) hence find 2 distinct ways of writing 65^2 as the sum a^2 + b^2, where a and b are integers and 0<a<b.

having trouble with part(ii)..
 

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,232
Gender
Undisclosed
HSC
N/A
w=(3+4i)/5
z=(5+12i)/13

wz=[(3+4i)(5+12i)]/65
=(-33+56i)/65

w(z cong)=[(3+4i)(5-12i)]/65
=(63-16i)/65

652=332+562
652=632+162
 
D

DkAssain

Guest
thanks :)

but is there an explanation on why the numerators of the solutions squared = to 65 ^2.
 

jet

Banned
Joined
Jan 4, 2007
Messages
3,148
Gender
Male
HSC
2009
Because |z| = |w| = 1
Then, |z||w| = 1
Then |zw| = 1
So then, calculating the mod of zw, (33^2 + 56^2/65^2) = 1 (by squaring)
<math>33^2 + 56^2 = 65^2</math>
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top