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Chapter 2: Problem 64

\(z^{2}=-z^{2}\)

Short Answer

Expert verified
z = 0

Step by step solution

01

Set Up the Equation

Start with the given equation: \[z^{2} = -z^{2}\]
02

Combine Like Terms

Move all terms to one side of the equation to set it to zero. Add \(z^{2}\) to both sides: \[z^{2} + z^{2} = 0\]
03

Simplify

Combine the terms on the left side: \[2z^{2} = 0\]
04

Solve for z

Divide both sides of the equation by 2: \[z^{2} = 0\] Now take the square root of both sides: \[z = 0\]

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Key Concepts

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

Equation Solving
Solving equations is a foundational skill in algebra. The goal is to find the value of the variable that makes the equation true.
For example, in the equation given: \(z^{2} = -z^{2}\), you want to isolate the variable z to find its value.
Remember these steps:
  • Move all terms to one side: This makes it easier to simplify the equation.
  • Simplify the equation: Combine like terms to get a simpler expression to work with.
  • Isolate the variable: Perform operations to solve for the unknown.
These steps can help you solve more complex equations as well.
Combining Like Terms
Combining like terms is a method to simplify equations. This means adding or subtracting terms with the same variable and exponent.
In our exercise, we combined the terms \(z^{2} + z^{2}\) to get \(2z^{2} = 0\).
When you have terms like \(3x\) and \(5x\), combining them gives \(8x\). Similarly, \(2y^{2} - y^{2}\) simplifies to \(y^{2}\).
This step is crucial before isolating the variable because it makes the equation simpler and easier to solve. Remember: always work with like terms to make the algebra easier.
Square Roots
Taking the square root of both sides is a common method for solving equations that involve squares.
In our exercise, after simplifying to \(z^{2} = 0\), we took the square root of both sides to find \(z\).
Remember: The square root function reverses the squaring operation. The square root of \(a^{2}\) is \(a\). For example, \(\sqrt{16} = 4\) because \(4^{2} = 16\).
  • When you take the square root of both sides, don’t forget about the \(\backslash \pm\) symbol for potential negative roots, but here it's not applicable since 0 has only one square root, which is 0.
  • Square roots help to simplify and solve equations where the variable is squared. It’s a powerful tool in algebra that opens the door to solving quadratic equations.

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Most popular questions from this chapter

In electricity we learn that the resistance of two resistors in series is \(R_{1}+R_{2}\) and the resistance of two resistors in parallel is \(\left(R_{1}^{-1}+R_{2}^{-1}\right)^{-1}\). Corresponding formulas hold for complex impedances. Find the impedance of \(Z_{1}\) and \(Z_{2}\) in series, and in parallel, given: (a) \(Z_{1}=2+3 i, \quad Z_{2}=1-5 i\) (b) \(Z_{1}=2 \sqrt{3} \angle 30^{\circ}, \quad Z_{2}=2 \angle 120^{\circ}\)

The three cube roots of \(+1\) are often called \(1, \omega\), and \(\omega^{2}\), Show that this is reasonable, that is, show that the cube roots of \(+1\) are \(+1\) and two other numbers, each of which is the square of the other.

Find and plot the complex conjugate of each number. \(\cos \pi-i \sin \pi\)

Find each of the following in the \(x+i y\) form. $$ \tanh \frac{3 \pi i}{4} $$

Describe geometrically the set of points in the complex plane satisfying the following equations. 51\. \(|z|=2\)
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