Answered

As long as its argument is restricted to [0,2), the polar form of a complex number is unique.

A. True
B. False



Answer :

To determine if the polar form of a complex number is unique when its argument is restricted to the interval [tex]\([0, 2\pi)\)[/tex], let's go through a step-by-step analysis.

1. Polar Form of a Complex Number:
The polar form of a complex number [tex]\( z \)[/tex] is expressed as:
[tex]\[ z = r \cdot e^{i\theta} \][/tex]
where:
- [tex]\( r \)[/tex] is the magnitude (absolute value) of the complex number.
- [tex]\( \theta \)[/tex] is the argument (angle) of the complex number.

2. Magnitude [tex]\( r \)[/tex]:
The magnitude [tex]\( r \)[/tex] of a complex number [tex]\( z = a + bi \)[/tex] is given by:
[tex]\[ r = \sqrt{a^2 + b^2} \][/tex]
The value of [tex]\( r \)[/tex] is always non-negative.

3. Argument [tex]\( \theta \)[/tex]:
The argument [tex]\( \theta \)[/tex] is the angle the complex number makes with the positive real axis, which can be calculated using the arctangent function:
[tex]\[ \theta = \arg(z) = \tan^{-1}\left(\frac{b}{a}\right) \][/tex]

4. Uniqueness Consideration:
For the polar form to be unique, we need to ensure that both [tex]\( r \)[/tex] and [tex]\( \theta \)[/tex] are uniquely determined. The magnitude [tex]\( r \)[/tex] is inherently unique and always positive. The critical part is whether the angle [tex]\( \theta \)[/tex] is unique.

5. Restriction on [tex]\(\theta\)[/tex]:
If we restrict [tex]\(\theta\)[/tex] to the interval [tex]\([0, 2\pi)\)[/tex], it means [tex]\(\theta\)[/tex] is confined within 0 (inclusive) and [tex]\(2\pi\)[/tex] (exclusive). This interval effectively maps each possible angle to a unique value within the range.

With this restriction in place, every complex number will have a distinct angle [tex]\(\theta\)[/tex] that fits within [tex]\([0, 2\pi)\)[/tex], ensuring that the polar representation [tex]\( r \cdot e^{i\theta} \)[/tex] is unique.

Therefore, the uniqueness condition is satisfied, leading us to the conclusion that the statement is True.

Answer: A. True