Basic complex numbers (1 Viewer)

[fullonoob]

Could you check if this is right please?

If z1 = 2 cis (5pi/6) and z2 = 3cis (2pi/3).
Write z1z2 in a) mod arg form
b) a + ib form

a) z1z2 = 2 cis (5pi/6)3cis (2pi/3)
= 6cis ( 3 pi / 2)
= 6 cis 270

b) a = rcos@ b = rsin@
Now what? No idea D:

thanks

 

[addikaye03]

Could you check if this is right please?

If z1 = 2 cis (5pi/6) and z2 = 3cis (2pi/3).
Write z1z2 in a) mod arg form
b) a + ib form

a) z1z2 = 2 cis (5pi/6)3cis (2pi/3)
= 6cis ( 3 pi / 2)
= 6 cis 270

b) a = rcos@ b = rsin@
Now what? No idea D:

thanks

z1 = 2 cis (5pi/6) and z2 = 3cis (2pi/3).

z1z2=6cis(5pi/6+2pi/3)=6cis(3pi/2)

b) it just wants it in a,b form (ie. not mod-arg) so you just gotta convert the above

6[cos(3pi/2)+isin(3pi/2)]

since in quad 3, cos 3pi/2=-cos(pi+@)=-cos(pi/2)=0

sin3pi/2=-sin(pi/2)=-1

therefore 6(0+i(-1))=0-6i