If z₁=5+5i and z₂=8(cos(3pi/7) +i sin(3pi/7)) then z₁z₂=

A. 40(cos(19pi/28)+i sin(19pi/28))
B. 40√2(cos(3pi/28)+i sin(3pi/28))
C. 8√5(cos(19pi/28)+i sin(19pi/28))
D. 40√2(cos(19pi/28)+i sin(19pi/28))

Question

If z₁=5+5i and z₂=8(cos(3pi/7) +i sin(3pi/7)) then z₁z₂=

A. 40(cos(19pi/28)+i sin(19pi/28))
B. 40√2(cos(3pi/28)+i sin(3pi/28))
C. 8√5(cos(19pi/28)+i sin(19pi/28))
D. 40√2(cos(19pi/28)+i sin(19pi/28))

anonymous · Answer

someone please help:/

inkyvoyd · Answer

You'l want to convert z1 to trig form (rcis theta) first.

anonymous · Answer

I have no idea how to do this problem what so ever so I need help :( a lot

inkyvoyd · Answer

and, @lgbasallote , I hve to go, so you are obligated to do this.

inkyvoyd · Answer

Or @ParthKohli

anonymous · Answer

is it c?

anonymous · Answer

@inkyvoyd @ParthKohli @lgbasallote

anonymous · Answer

:(

anonymous · Answer

tHats pretty simple just multiplythe complex numbers ..
the answer is a
On multiplying u get 40((cos 3pi/7 -  sin 3pi/7) +i(cos3pi/7 + sin 3pi/7))
You can see the form of cos a cos b - sin a sin b
And sin a cos. b +Sin b cos a
On multiplying the terms by 1/sqr root 2