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Suppose \(z_{1}\) and \(z_{2}\) are complex numbers. What can be said about \(z_{1}\) or \(z_{2}\) if \(z_{1} z_{2}=0 ?\)

Short Answer

Expert verified
If \(z_1 z_2 = 0\), then \(z_1 = 0\) or \(z_2 = 0\).

Step by step solution

01

Understand the Property of Complex Multiplication

In the field of complex numbers, if the product of two complex numbers is zero, i.e., \(z_1 z_2 = 0\), it implies that at least one of the numbers must be zero. This property stems from the absence of zero divisors in the complex number system, similar to real numbers.
02

Apply the Property to the Given Problem

Given the equation \(z_1 z_2 = 0\), based on the property from Step 1, we can conclude that either \(z_1 = 0\) or \(z_2 = 0\). This implies that one or both of the complex numbers must be zero.
03

Example Verification

Consider \(z_1 = 0\) and \(z_2\) any non-zero complex number. The product \(z_1 z_2 = 0 \times z_2 = 0\). Similarly, if \(z_1\) is non-zero and \(z_2 = 0\), then \(z_1 z_2 = z_1 \times 0 = 0\). This verifies our solution logically.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Multiplication
Complex multiplication involves multiplying two complex numbers, each composed of a real and an imaginary part. A complex number is generally represented as \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit satisfying \( i^2 = -1 \). When you multiply two complex numbers \( z_1 = a + bi \) and \( z_2 = c + di \), the distributive law is applied:
  • Multiply the real parts to get \( ac \).
  • Multiply the real part of the first with the imaginary part of the second to get \( adi \).
  • Multiply the imaginary part of the first with the real part of the second to get \( bci \).
  • Finally, multiply the imaginary parts to get \( bdi^2 = -bd \) (since \( i^2 = -1 \)).
Combine the contributions to get the resultant complex number: \[ z_1 z_2 = (ac - bd) + (ad + bc)i \].
The multiplication of complex numbers is straightforward, but the absence of zero divisors plays a crucial role. As seen in the problem set up, if the product \(z_1 z_2 = 0\), understanding this multiplication process helps determine that either \(z_1\) or \(z_2\) must itself be zero, revealing insights into how complex numbers operate.
Zero Divisors
In the realm of complex numbers, zero divisors are non-existent. A zero divisor is a non-zero element in a ring that can multiply with another non-zero element to give zero. However, in the complex number system, if the product \( z_1 z_2 = 0 \), one, or both, numbers must be zero. This is akin to the real number system where the absence of such elements illustrates a distinct property known as the 'cancelation law.'
The absence of zero divisors in complex numbers is critical. It guarantees that solutions to equations like \( z_1 z_2 = 0 \) are straightforward, ensuring that if neither number is zero, their product can never be zero. This is why, based on the solution provided in the original exercise, we conclude that if \( z_1 z_2 = 0 \), at least one number among \( z_1 \) and \( z_2 \) needs to be zero. This absence of zero divisors is a core pillar upholding the integrity of operations and solutions within the complex number system.
Complex Number System
The complex number system extends the real number system by including imaginary numbers. It is a rich field of mathematics where numbers are expressed in the form \( z = a + bi \).
  • The real part is \( a \).
  • The imaginary part is \( b \) with the imaginary unit \( i \).
  • Complex numbers are crucial in solving polynomial equations where real number solutions fall short.

The geometry of complex numbers allows them to be represented on a plane, called the complex plane, where the x-axis represents the real part and the y-axis the imaginary part.
The complex number system is fascinating because it displays properties akin to real numbers while opening possibilities that real numbers alone cannot describe. It encompasses operations of addition, subtraction, multiplication, and division, adhering to algebraic rules.
The lack of zero divisors within this system parallels that of real numbers, accentuating a fundamental property where the product of two complex numbers is zero only if one is zero itself. This characteristic is highlighted in the exercise, where the absence of zero divisors directly leads to the conclusion regarding \( z_1 \) and \( z_2 \). Understanding complex numbers enhances comprehension in numerous fields such as engineering, physics, and computer science, providing solutions to problems inefficiently tackled by real numbers alone.

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Most popular questions from this chapter

Cubic Formula In this project you are asked to investigate the solution of a cubic polynomial equation by means of a formula using radicals, that is, a combination of square roots and cube roots of expressions involving the coefficients. (a) To solve a general cubic equation \(z^{3}+a z^{2}+b z+c=0\) it is sufficient to solve a depressed cubic equation \(x^{3}=m x+n\) since the general cubic equation can be reduced to this special case by eliminating the term \(a z^{2}\). Verify this by means of the substitution \(z=x-a / 3\) and identify \(m\) and \(n\). (b) Use the procedure outlined in part (a) to find the depressed cubic equation for \(z^{3}+3 z^{2}-3 z-9=0\) (c) A solution of \(x^{3}=m x+n\) is given by $$ x=\left[\frac{n}{2}+\left(\frac{n^{2}}{4}-\frac{m^{3}}{27}\right)^{1 / 2}\right]^{1 / 3}+\left[\frac{n}{2}-\left(\frac{n^{2}}{4}-\frac{m^{3}}{27}\right)^{1 / 2}\right]^{1 / 3}. $$ Use this formula to solve the depressed cubic equation found in part (b). (d) Graph the polynomial \(z^{3}+3 z^{2}-3 z-9\) and the polynomial from the depressed cubic equation in part (b); then estimate the \(x\) -intercepts from the graphs. (e) Compare your results from part (d) with the solutions found in part (c). Resolve any apparent differences. Find the three solutions of \(z^{3}+3 z^{2}-\) \(3 z-9=0\) (f) Do some additional reading to find geometrically motivated proofs (using a square and a cube) to derive the quadratic formula and the formula given in part (c) for the solution of the depressed cubic equation. Why is the name quadratic formula used when the prefix quad stems from the Latin word for the number four?

Use a CAS as an aid in factoring the given quadratic polynomial. $$ z^{2}-\sqrt{3} z-i $$

Sketch the set of points in the complex plane satisfying the given inequality. $$ -\pi<\arg (z)<\pi / 2 $$

Sketch the set of points in the complex plane satisfying the given inequality. $$ 0 \leq \arg (z) \leq \pi / 6 $$

Express the given complex number in the exponential form \(z=r e^{i \theta}\). $$ -4-4 i $$

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