Math Problem Statement

If z = (-1 - j)^2 / 2, write z in three forms and solve: a) b) z^2 - 1, c) find the cube roots of z.

Solution

The problem you've uploaded seems to be related to a complex number z=(1j)22z = \frac{(-1 - j)^2}{2}, where the goal is to represent zz in three forms, as well as calculate its powers and roots.

Let's break it down step by step:

Part a) Represent zz in three forms (likely Cartesian, Polar, and Exponential form).

Part b) Compute z21z^2 - 1.

Part c) Find the cube roots of zz.

I will proceed to solve these now. Let's start with part a.

Step 1: Simplify the given expression z=(1j)22z = \frac{(-1 - j)^2}{2}.

First, let’s simplify (1j)2(-1 - j)^2, where jj is the imaginary unit.

(1j)2=(1)2+2(1)(j)+(j)2=1+2j1=2j(-1 - j)^2 = (-1)^2 + 2(-1)(-j) + (-j)^2 = 1 + 2j - 1 = 2j

So, the expression for zz becomes: z=2j2=jz = \frac{2j}{2} = j

Thus, z=jz = j, a purely imaginary number.

Part a) Now, we write zz in the three standard forms:

  1. Cartesian form (a + bi): z=0+1jz = 0 + 1j

  2. Polar form (r * (cos(θ) + i sin(θ))): The magnitude rr is: r=z=j=1r = |z| = |j| = 1 The argument θ\theta is π2\frac{\pi}{2}, since zz is along the positive imaginary axis.

    Thus, the polar form is: z=1(cos(π2)+isin(π2))z = 1 (\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}))

  3. Exponential form (r * e^(iθ)): Using the same values for rr and θ\theta: z=eiπ2z = e^{i \frac{\pi}{2}}

Part b) Now, let's calculate z21z^2 - 1:

We know z=jz = j, so: z2=j2=1z^2 = j^2 = -1 Thus: z21=11=2z^2 - 1 = -1 - 1 = -2

Part c) Finding the cube roots of zz:

To find the cube roots of zz, we use the polar form of z=eiπ2z = e^{i\frac{\pi}{2}}.

The cube roots of a complex number are given by: zk=r1/3ei(θ+2kπ3),k=0,1,2z_k = r^{1/3} e^{i(\frac{\theta + 2k\pi}{3})}, \quad k = 0, 1, 2

For z=eiπ2z = e^{i\frac{\pi}{2}}, r=1r = 1 and θ=π2\theta = \frac{\pi}{2}, so the cube roots are: zk=ei(π/6+2kπ3),k=0,1,2z_k = e^{i(\frac{\pi/6 + 2k\pi}{3})}, \quad k = 0, 1, 2

Thus, the cube roots are:

  1. z0=eiπ6z_0 = e^{i\frac{\pi}{6}}
  2. z1=ei5π6z_1 = e^{i\frac{5\pi}{6}}
  3. z2=ei3π2z_2 = e^{i\frac{3\pi}{2}}

These are the three cube roots of zz.


Would you like a more detailed explanation of any of these steps? Or if you have further questions, feel free to ask!

Here are some related questions:

  1. What is the difference between the Cartesian, Polar, and Exponential forms of complex numbers?
  2. How can you calculate the roots of a complex number using its polar form?
  3. How do you perform complex number exponentiation?
  4. What is the significance of the argument in polar and exponential forms?
  5. How would you compute higher powers or roots of complex numbers?

Tip: When working with complex numbers in polar or exponential form, always remember to convert between forms carefully, especially when finding powers or roots.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Form
Exponential Form
Roots of Complex Numbers

Formulas

Cartesian Form: z = a + bi
Polar Form: z = r (cos(θ) + i sin(θ))
Exponential Form: z = r e^(iθ)
Roots of Complex Numbers: z_k = r^(1/n) e^(i(θ + 2kπ)/n), k = 0, 1, ..., n-1

Theorems

De Moivre's Theorem

Suitable Grade Level

Grades 11-12