Polar Form of Complex Numbers: Solving z z2 - 1
In mathematics, particularly in the field of complex analysis, the polar form of a complex number z is a powerful and elegant representation that provides a deeper understanding of the number's properties. Let's explore the polar form of the complex number z z2 - 1 and how to solve it in polar form.
Introduction to Complex Numbers
A complex number is a number of the form a bi, where a and b are real numbers, and i is the imaginary unit with the property i2 -1. The polar form of a complex number provides an alternative representation that uses the modulus (or magnitude) and angle (or argument) of the number. This form is particularly useful for operations involving multiplication and division of complex numbers.
Understanding Polar Form
The polar form of a complex number z is given by:
z r ? eit
Where r is the modulus (distance from the origin) of z, and θ (theta) is the argument of z (the angle that z makes with the positive real axis).
In the polar form, we can express z as:
z r ? eitheta
For the equation z z2 - 1, let's transform it into polar form to solve for z.
Solving the Equation in Polar Form
Let z r ? eitheta. Substituting z into the equation, we get:
r ? eitheta (r ? eitheta)2 - 1
This can be rewritten as:
r ? eitheta r2 ? e2itheta - 1
For this equation to hold, we need to consider two cases: the real and imaginary parts.
Case 1: Modulus (r)
The magnitude of the left side is r, and the magnitude of the right side is |r2 ? e2itheta - 1|. For the equation to hold, the magnitudes must be equal:
r |r2 ? e2itheta - 1|
Considering the properties of the modulus, we can further analyze the possible values of r.
Case 2: Argument (θ)
For the equation to hold, the arguments must also be equal:
θ 2θ 2kπ, or θ 2kπ
Where k is an integer. This implies that the argument of the complex number is either 0 or an integral multiple of 2π.
Finding the Solutions
Now, to find the specific solutions, we need to solve the equation z2 - 1 0:
z2 1
taking the square root of both sides, we get:
z 1 or z -1
Expressing these solutions in polar form:
z 1 1 ? ei0, and z -1 1 ? eiπ
However, to find all the solutions, we can use the quadratic formula:
z1,2 (1 ± √(1 - 4)) / 2 (1 ± √(-3)) / 2 (1 ± isqrt(3)) / 2
Thus, we have:
z1 (1 / 2) (sqrt(3) / 2) i, and z2 (1 / 2) - (sqrt(3) / 2) i
Converting to Polar Form
Now, we convert these solutions to polar form:
z1 (1 / 2) (sqrt(3) / 2) i r1 ? eitheta1, where r1 1 and theta1 pi/3
z2 (1 / 2) - (sqrt(3) / 2) i r2 ? eitheta2, where r2 1 and theta2 -pi/3
Conclusion
The polar form of the complex number z in the equation z z2 - 1 is a powerful tool for understanding and solving problems involving complex numbers. By converting the solutions to polar form, we can gain deeper insights into the properties of these numbers and their relationships.