Every nonzero complex number has exactly​ 𝑛 distinct 𝑛th​ root(s).Group of answer choicesTrueFalse

Step 1: Break Down the Problem

The statement claims that every nonzero complex number has exactly $n$ distinct $n$th roots. We need to analyze whether this statement is true or false.

Step 2: Relevant Concepts

In complex analysis, for any nonzero complex number $z$ in polar form, given by: $z = re^{i\theta}$ where $r$ is the magnitude and $\theta$ is the argument of the complex number, the $n$th roots can be expressed as: $z_k = r^{1/n}\left(e^{i\theta/n}e^{2\pi ik/n}\right) \quad \text{for } k = 0, 1, 2, \ldots, n-1$ This means that there are $n$ distinct values (roots) for each $k$.

Step 3: Analysis and Detail

1. Magnitude: The magnitude of each root $z_k$ is $r^{1/n}$.
2. Argument: The arguments of the $n$th roots are given by: $\frac{\theta + 2\pi k}{n} \quad \text{for } k = 0, 1, 2, \ldots, n-1$ Since $k$ takes on $n$ distinct values, it results in $n$ distinct angles.

The roots are distinct because:

• The angles $\frac{\theta + 2\pi k}{n}$ for different $k$ values differ by $\frac{2\pi}{n}$, which means they map to different points on the complex plane, specifically distributed evenly around the circle of radius $r^{1/n}$.

Step 4: Verify and Summarize

Since each nonzero complex number consistently yields $n$ unique distinct $n$th roots, we conclude that the statement is indeed valid.

Final Answer

True: Every nonzero complex number has exactly $n$ distinct $n$th roots.