Why can z^3 = -27 be written as r^3(cos(3(theta)+i.sin(3(theta))?

Posted on 2011-02-21
Last Modified: 2012-05-11
I'm working with complex numbers and reading through a solution,

It begins like this - In polar form, -27 = 27 (cos(pi) + i sin(pi))

I get that, but then it says

If z = r(cos(theta)+i sin(theta), then the equation z^3 = -27 can be written as

r^3(cos(3(theta)+i.sin(3(theta)) = 27(cos(pi) + i sin(pi))

I don't see where the left hand side came from??

Question by:purplesoup
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LVL 32

Expert Comment

ID: 34942649
It comes from Euler's formula -'s_formula

e^(i theta) == cos(theta) + i sin(theta)

LVL 32

Accepted Solution

phoffric earned 250 total points
ID: 34942657
z  = r( cos(theta) + i sin(theta) )
 z³ = r³( cos(theta) + i sin(theta) )³

Convert to e^p
raise to the third power
and convert back again to polar form
LVL 18

Assisted Solution

deighton earned 250 total points
ID: 34942760
lets see

z1 = r1(cos(A)+i sin(A))
z2 = r2(cos(B)+i sin(B))

z1 x z2 = r1(cos(A)+i sin(A)) r2(cos(B)+i sin(B))

= r1 r2 [cos(A) cos(B) + i sin(A) cos(B) + i cos(A) sin(B) - sin(A) sin(B)]

now since we know where we want to go, let us recall the double angle identities

sin(A + B) = sin(A)cos(B) + sin(B) cos(A)
cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

using these

 r1 r2 [cos(A) cos(B) + i sin(A) cos(B) + i cos(A) sin(B) - sin(A) sin(B)]

=  r1 r2 [cos(A) cos(B)  - sin(A) sin(B) +  i sin(A) cos(B) + i cos(A) sin(B))]

=  r1 r2 [cos(A+B) +  i sin(A + B)]


so in general, the argument angle is always added when multiplying two complex numbers.  So if a number is cubed, the angle is tripled

if the angle exceeds 360 degrees (or 2 pi radians as you will have to come to know it) the argument can be reduced back to the corresponding value 0<= arg <= 2pi


Author Closing Comment

ID: 34942901
That's great - thanks!

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