Solve the equation z^5=-1 giving the solutions in polar form
I've got z^5 =1 but just not sure as its a negative
to get -1 + 0 i
z^5 = 1 cos 180 + i sin 180
That should get you started
let z^5 = rcosθ + rsinθ i
= -1 + 0
we have r = 1
θ = 180°
so z^5 = 1(cos180° + isin180°)
by de Moivre's theorme
z = 1^(1/5)(cos 180/5 + i sin180/5)
= 1(cos 36° + isin36°) or cis36°
repeating at 360/5 or 72°
z = cis 36°
z = cis108°
z = cis180° -----> there is our -1
z= cis252°
z = cis324°
To solve the equation z^5 = -1, you first need to express -1 in polar form. The polar form of -1 is given by -1 = e^(iπ+2kπ), where k is an integer.
Now, let's find the solutions by raising z to the power of 1/5. We have:
z = (-1)^(1/5) = (e^(iπ+2kπ))^(1/5)
To simplify this expression, we use the properties of exponents to combine the exponents inside the parentheses:
z = e^((iπ+2kπ)(1/5))
Now, we simplify the exponent further:
z = e^(iπ/5 + 2kπ/5)
This is the polar form of z. In polar form, a complex number z is given by z = re^(iθ), where r is the modulus (magnitude) of z, and θ is the argument (angle) of z.
Comparing the form of z = e^(iπ/5 + 2kπ/5) with z = re^(iθ), we can identify the modulus and argument:
Modulus: r = 1
Argument: θ = π/5 + 2kπ/5, where k is an integer
Therefore, the solutions to the equation z^5 = -1 in polar form are:
z₁ = 1*e^(iπ/5)
z₂ = 1*e^(i3π/5)
z₃ = 1*e^(i5π/5)
z₄ = 1*e^(i7π/5)
z₅ = 1*e^(i9π/5)
These are the five complex solutions in polar form.