Problem 16 - Precalculus

Question

Express each complex number in polar form. $$-\sqrt{3}+i$$

Step-by-Step Solution

Answer

The polar form is \( 2 \operatorname{cis}\left(\frac{5\pi}{6}\right) \).

Step 1: Convert to Polar Coordinates

Recall that a complex number can be expressed in the form \( a + bi \), which corresponds to a point \( (a, b) \) in the complex plane. Here, \( a = -\sqrt{3} \) and \( b = 1 \). To convert this to polar form, we use the polar coordinates \( r \) (magnitude) and \( \theta \) (angle).

Step 2: Compute the Magnitude

The magnitude \( r \) of a complex number \( a + bi \) is calculated as \( r = \sqrt{a^2 + b^2} \). Substitute \( a = -\sqrt{3} \) and \( b = 1 \):\[ r = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2. \]This simplifies to \( r = 2 \).

Step 3: Determine the Angle

The angle \( \theta \), or argument, can be found using the formula \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \). Here it is \( \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) \). This is equivalent to an angle where the tangent is negative in the second quadrant, making \( \theta = \frac{5\pi}{6} \).

Step 4: Write in Polar Form

Once you have \( r = 2 \) and \( \theta = \frac{5\pi}{6} \), the polar form of the complex number \( -\sqrt{3}+i \) is expressed as \( 2 \operatorname{cis}\left(\frac{5\pi}{6}\right) \), which is equivalent to \( 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) \).

Key Concepts

Understanding Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part, typically expressed as \( a + bi \). Here, \( a \) is the real component, and \( bi \) is the imaginary component, where \( i \) is the imaginary unit and equal to \( \sqrt{-1} \). In the case of our exercise, the complex number is \( -\sqrt{3} + i \).

This can be thought of as a point on the complex plane, where the horizontal axis represents the real part, and the vertical axis represents the imaginary part. It is essential to understand that complex numbers provide a valuable way to represent two-dimensional quantities.

• The real part \( -\sqrt{3} \) plots to the left of the origin on the horizontal axis.
• The imaginary part \( i \) moves upwards on the vertical axis, above the origin.

Converting complex numbers into polar form allows for easier multiplicative operations, and this format is based on magnitude and angle.

Finding Magnitude of Complex Numbers

The magnitude, often denoted as \( r \), is akin to the distance from the origin to the point representing the complex number in the complex plane. This is calculated with the formula \( r = \sqrt{a^2 + b^2} \).

For our given number, \( a = -\sqrt{3} \) and \( b = 1 \). Substituting these values into the formula gives:
\[r = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2.\]

Thus, the magnitude \( r \) is 2.

• This tells us the point \(-\sqrt{3} + i\) is 2 units away from the origin.
• The magnitude provides us with the scaling factor for the direction determined by the angle.

This magnitude is an essential part of expressing the complex number in polar coordinates.

Angle in the Second Quadrant

Identifying the correct angle in the polar coordinate system is crucial. The quadrant tells us about the direction of the point from the origin. Since \( a = -\sqrt{3} \) and \( b = 1 \), our complex number is located in the second quadrant.

To determine the angle, \( \theta \), we use the tangent inverse formula:
\( \theta = \tan^{-1}\left(\frac{b}{a}\right) \) or \( \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) \).

• In the second quadrant, the tangent is negative.
• The specific angle that satisfies \( \tan(\theta) = -\frac{1}{\sqrt{3}} \) is \( \frac{5\pi}{6} \).

This angle \( \frac{5\pi}{6} \) situates the complex number correctly in the second quadrant, completing the description in polar form. Understanding this allows for accurate calculations in many practical scenarios.