Problem 109
Question
Determine whether each statement makes sense or does not make sense, and explain your reasoning. My work with complex numbers verified that the only possible cube root of 8 is 2
Step-by-Step Solution
Verified Answer
The statement does not make sense. While 2 is indeed a cube root of 8 in the real number system, there are other solutions in the complex number system.
1Step 1: Understanding Cube Roots
In the real number system, a cube root of a number \(n\) is a value \(a\) such that \(a^3 = n\). So, in the real number system, the cube root of 8 is indeed 2.
2Step 2: Dealing with Complex Numbers
Understand that in the complex number system, a number can actually have more than one cube root. This is because complex numbers involve both real and imaginary components, and “cubing” a complex number involves rotating it in the complex plane. The cube roots of a number are evenly distributed around a circle in the complex plane and are 120 degrees apart.
3Step 3: Conclusion and Reasoning
In the context of complex numbers, the statement that 'the only possible cube root of 8 is 2' is incorrect. In the complex number system, 8 actually has three cube roots. One of these is the real number 2, and the other two are complex numbers: \(-1 + \sqrt{3}i\) and \(-1 - \sqrt{3}i\). Therefore, the statement in the problem does not make sense, because it inaccurately claims that there is only one cube root of 8.
Other exercises in this chapter
Problem 107
Prove that the distance, \(d\), between two points with polar coordinates \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) is $$ d=\sqr
View solution Problem 108
Determine whether each statement makes sense or does not make sense, and explain your reasoning. The proof of the formula for the product of two complex numbers
View solution Problem 109
Graph \(y=2 \sin \frac{1}{2} x .\) Then use the graph to obtain the graph of \(y=2 \csc \frac{1}{2} x .\) (Section \(5.6,\) Example 4 )
View solution Problem 110
Prove the rule for finding the quotient of two complex numbers in polar form. Begin the proof as follows, using the conjugate of the denominator's second factor
View solution