Question

Find z*w. z=5(cos35+isin35) and w=2(cos40+isin40).

Answer 1

we have a formula to find it out. It says
z*w = number in the front of z* number in the front of w *(cos (angle of z + angle of w) + i sin (angle of z + angle of w))
show my your understanding

Answer 2

Would it be 10(cos75+isin75)? (75 being in degrees)

Answer 3

Proud of you!! :)

Answer 4

Thanks, could you help me with writing in a+bi form?

Answer 5

we have formula to do it, too
z= r , theta
so, from yours r = 10, theta = 75
the formula to write the complex form is r = x +iy
where x = r cos (theta)
y = r sin (theta)
show me your work

Answer 6

I take the result as z , it is not original z.

Answer 7

The expression I need to write in a+bi form is (sqrt3 + i)^5.

Answer 8

ohhh, new problem?

Answer 9

Yes, sorry

Answer 10

let say the number is z = (sqrt3 +i)^5

Answer 11

is it not just open the bracket? you have the form (a + bi )already? where a = sqrt 3, and b =1??

Answer 12

but I don't think we have to do it. it forms the form you need

Answer 13

That's why I am confused. The 4 answers to choose from are
16-16i(sqrt3)
-16(sqrt3)+16i
16(sqrt3)+16i
-16(sqrt3)-16i

Answer 14

one way to do, just (sqrt 3 +i)(sqrt3+i)(sqrt3+i)(sqrt3+i)(sqrt3+i)
and open the first 2 terms = 3 +2sqrt3*i +i^2 = 2+ 2sqrt3*i
open the next 2 terms, get the same
so far, we have (2+2sqrt3*i)(2+2sqrt3 *i)(sqrt3 +i)
next, the first 2 terms of the new one is.....
if you know how to use Pascal triangle , the problem will be easier to solve
but if you don't , just count your fingers as what I did, hehehe