Question

let z1= 2 ( cos(pi/5) )+ i sin (pi/5) ) and z2= 8 (cos (7pi/8) + i sin (7pi/6) )

calculate :z1/z2

Answer 1

please check it is z2= 8 (cos (7pi/8) + i sin (7pi/6) ) . that is pi/8 and pi/6

Answer 2

This can be done by realising that when you divide two complex numbers, you subtract arguments and divide the magnitudes.

So: Arg(z1/z2)=arg(z1)-arg(z2)
And: |z1/z2|=|z1|/|z2|

Answer 3

Now the argument of z1 is pi/5, that of z2 is 7pi/8
Also: |z1|=2, and |z2|=8

Answer 4

my bad its 8cos7pi/6

Answer 5

cosx + i sinx = e^ix

z1= 2 ( cos(pi/5) )+ i sin (pi/5) ) = 2 e^i(pi/5)
and
z2= 8 (cos (7pi/6) + i sin (7pi/6 ) = 8 e^(7pi/6)

Answer 6

So the argument of Z1/z2 is pi/5-7pi/6=-29pi/30
|z1/z2| = 2/8=1/4

Answer 7

Putting it together: z1/z2=1/4(cos(-29pi/30+i sin(-29pi/30))

Answer 8

for z1/z2 = 2 e^i(pi/5)/8 e^(7pi/6) = 1/4 e^ i( pi/5 - 29/5) = 1/4 e^i(-29pi/30)
which is equal to 1/4(cos(-29pi/30+i sin(-29pi/30))

Answer 9

thats the answer?

Answer 10

OK, @cinmin, after all this (mostly duplicate ;) ) explanations of @ritez and myself, do you understand?

Answer 11

basically we just combine them right?

Answer 12

We combine them in this way: we subtract the arguments (angles) and we divide the magnitudes (or absolute values, or moduli, or whatever you call it), yes.

Answer 13

as ritez said: 1/4(cos(-29pi/30+i sin(-29pi/30)) which is 1/4e^i(-29pi/30)

Answer 14

oh okay it makes more sense now. Thank you all!

Answer 15

YW!