I want to solve z^5=-1
But all the examples I've seen use positive numbers. The -1 is throwing me off somewhat.
Most of the examples I've worked with would be solved using the polar form I.e
Z^5(cos5+i sin5)=1(cospi +i sinpi)
But as I said negative numbers I'm finding confusing.
huh? negatives are just numbers, like positives.
z^5 = -1 = 1 cisπ
z = 1^(1/5) cis(π/5)
but since cisπ = cis3π = cis5π, etc., to get all the values from 0 to 2π,
z = 1 cis(π/5 k) where k = 1,3,5,7,9
Reiny went over this with you. There are 5 5th-roots of -1. 4 of them are complex, and one of them is real: -1.
This question must be a popular one.
Did it yesterday
http://www.jiskha.com/display.cgi?id=1461490844
To solve the equation z^5 = -1, it's helpful to express -1 in polar form, just like you mentioned.
The polar form of a complex number is given by z = r(cosθ + i sinθ), where r is the magnitude (distance from the origin to the point) and θ is the angle (in radians) that the point makes with the positive real axis.
For -1, we can write it as -1 = 1(cosπ + i sinπ).
Now, we can compare z^5 = -1 with z = 1(cosπ + i sinπ):
(z^5)^2 = (-1)^2
z^10 = 1.
This means that we need to find the complex numbers z whose 10th power is equal to 1.
To obtain the solutions, we can express 1 in polar form: 1 = 1(cos0 + i sin0).
Now consider the general form of z:
z = r(cosθ + i sinθ)
If we raise z to the 10th power, we get:
(z^10) = (r(cosθ + i sinθ))^10
= r^10 (cos(10θ) + i sin(10θ)).
For z^10 to equal 1, we have:
r^10 (cos(10θ) + i sin(10θ)) = 1(cos0 + i sin0).
Since the magnitudes should be equal, we have r^10 = 1. This gives us r = 1.
To find the possible values of θ, we use the fact that the cosine and sine functions have periodicity of 2π. Therefore, we can write:
10θ = 2nπ, where n is an integer.
This gives us:
θ = (2nπ)/10
= (nπ)/5, where n is an integer.
So, we have a total of 10 possible values for θ, given by:
θ = (0π)/5, (1π)/5, (2π)/5, (3π)/5, (4π)/5, (5π)/5, (6π)/5, (7π)/5, (8π)/5, (9π)/5.
Substituting these values back into the polar form expression for z, we get 10 solutions for z that satisfy z^5 = -1.