Problem 22 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-46: Express the complex number in trigonometric form with \(0 \leq \theta<2 \pi\). $$ \sqrt{3}+i $$

Step-by-Step Solution

Answer

The complex number \( \sqrt{3} + i \) in trigonometric form is \( 2(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}) \).

Step 1: Identify Real and Imaginary Parts

The complex number given is \( \sqrt{3} + i \). Here, the real part \( a \) is \( \sqrt{3} \) and the imaginary part \( b \) is 1.

Step 2: Calculate the Magnitude

The magnitude \( r \) of the complex number is found using the formula \( r = \sqrt{a^2 + b^2} \). So, \( r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \).

Step 3: Find the Argument

The argument \( \theta \) is found using \( \tan \theta = \frac{b}{a} \). Here, \( \tan \theta = \frac{1}{\sqrt{3}} \). This corresponds to \( \theta = \frac{\pi}{6} \) because \( \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \).

Step 4: Express in Trigonometric Form

Now that we have \( r = 2 \) and \( \theta = \frac{\pi}{6} \), the trigonometric form is \( 2 (\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) \).

Key Concepts

Complex Numbers

Complex numbers are numbers that include a real part and an imaginary part. The imaginary part is denoted with an 'i', which represents the square root of -1. This might seem complex at first, but it is just another way of expanding numbers as we move beyond the number line.
Think of a complex number as any number written in the form \( a + bi \), where

• \( a \) is the real part and
• \( b \) is the imaginary part.

For example, in the complex number \( \sqrt{3} + i \), \( \sqrt{3} \) is the real part, and \( 1 \) is the imaginary part. Complex numbers are useful in advanced mathematics, engineering, and physics because they allow calculations that cannot be done with real numbers alone, such as those involving rotation and oscillation.

Magnitude of Complex Numbers

The magnitude of a complex number gives us a sense of its size or distance from the origin when plotted on the complex plane. It is similar to calculating the length of a line segment in a coordinate plane.
We can find the magnitude, usually denoted by \( r \), by using the formula:

• \( r = \sqrt{a^2 + b^2} \)

Where \( a \) is the real part and \( b \) is the imaginary part of the complex number \( a + bi \). In simpler terms, it's like using the Pythagorean theorem, where you find the hypotenuse of a right triangle.
For the number \( \sqrt{3} + i \), plug \( a = \sqrt{3} \) and \( b = 1 \) into the formula to get:

• \( r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \)

Thus, the magnitude is \( 2 \).

Argument of Complex Numbers

The argument of a complex number is like the angle it forms with the positive real axis in the complex plane. It helps determine the direction or "rotation" of the complex number from the origin.
To find the argument \( \theta \), you use the formula:

• \( \tan \theta = \frac{b}{a} \)

In our example, where \( \sqrt{3} + i \), \( a = \sqrt{3} \) and \( b = 1 \), the argument \( \theta \) is found by solving:

• \( \tan \theta = \frac{1}{\sqrt{3}} \)

We know that the value \( \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \), hence \( \theta = \frac{\pi}{6} \).
It's important to keep \( \theta \) within the range \( 0 \leq \theta < 2\pi \), which corresponds to one full rotation in a circle.

Polar Coordinates

Polar coordinates offer a way of locating a point in a plane through a distance and an angle, rather than traditional x/y axes. When expressing a complex number in polar coordinates, you focus on its magnitude and argument.
The coordinate system is defined as \( (r, \theta) \), where:

• \( r \) is the magnitude of the complex number, and
• \( \theta \) is the argument.

To convert the complex number \( a + bi \) into trigonometric form, you represent it as

• \( r ( \cos \theta + i \sin \theta ) \)

For \( \sqrt{3} + i \), we found:

• \( r = 2 \)
• \( \theta = \frac{\pi}{6} \)

The polar/trigonometric form then becomes \( 2 ( \cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}) ) \)
This showcases how simples circles and triangles intermingle in complex number calculations, making them easily workable in various mathematical solutions and applications.