Problem 134 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the square roots of the complex number. $$2+2 i$$

Step-by-Step Solution

Answer

The square roots of the complex number \(2+2i\) are \(\sqrt{2}(\cos(\frac{\pi}{8}) + i \sin(\frac{\pi}{8}))\) and \(\sqrt{2}(-\cos(\frac{\pi}{8}) - i \sin(\frac{\pi}{8}))\).

Step 1: Calculate the modulus (r)

To find \(r\), we use the formula \(r = \sqrt{a^2 + b^2}\), where \(a\) is the real part of the complex number and \(b\) is the image. Therefore, \(r = \sqrt{(2)^2 + (2)^2} = 2\sqrt{2}\).

Step 2: Compute the argument (θ)

The argument \(θ\) is given by \(\theta = \arctan \frac{b}{a}\), where \(a\) is the real part and \(b\) is the imaginary part of the complex number. Therefore, \(\theta = \arctan \frac{2}{2} = \arctan 1 = \frac{\pi}{4}\)

Step 3: Insert r and θ into the square root formula

Now we can calculate \(\sqrt{z}\) using the formula \(\sqrt{z} = \sqrt{r}(\cos(\theta/2) + i \sin(\theta/2))\). We get \(\sqrt{z} = \sqrt{2\sqrt{2}}(\cos(\frac{\pi}{8}) + i \sin(\frac{\pi}{8}))\). One root is positive and another root is negative, thus giving \(\sqrt{2}(\cos(\frac{\pi}{8}) + i \sin(\frac{\pi}{8}))\) and \(\sqrt{2}(-\cos(\frac{\pi}{8}) - i \sin(\frac{\pi}{8}))\).

Key Concepts

Modulus of a Complex Number

Understanding the modulus of a complex number is crucial when working with complex arithmetic. The modulus refers to the distance of a complex number from the origin in the complex plane. It is denoted by \(|z|\) and is calculated 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 \(z = a + bi\).
For example, let's take the complex number \(2 + 2i\). Applying the formula, we get the modulus \(r = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\). This value represents the length of the vector from the origin to the point \((2, 2)\) on the complex plane, providing a geometric interpretation of the complex number's magnitude.

Argument of a Complex Number

The argument of a complex number is the angle it makes with the positive direction of the real axis, often measured in radians. Symbolically, given a complex number \(z = a + bi\), its argument is denoted by \(\theta\) and can be found using the formula \(\theta = \arctan \frac{b}{a}\).

Finding the Argument

For the complex number \(2+2i\), one would compute \(\theta = \arctan \frac{2}{2} = \arctan(1) = \frac{\pi}{4}\). This angle is the direction of the vector representing our complex number in the complex plane. Importantly, the argument is not unique – adding multiples of \(2\pi\) to an argument gives another valid argument for the same complex number because angles in the trigonometric circle are periodic. Thus, identifying the argument helps in expressing the complex number in polar or trigonometric form.

Trigonometric Form of Complex Numbers

The trigonometric form of complex numbers is a powerful concept that facilitates various operations, such as multiplication, division, and finding roots. A complex number can be expressed as \(z = r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the modulus and \(\theta\) is the argument. This form highlights the polar coordinates of the complex number on the complex plane.

Applying Trigonometric Form to Square Roots

To find square roots of a complex number, we first convert it into its trigonometric form and then apply the formula for the square root of a complex number which is \(\sqrt{z} = \sqrt{r}(\cos(\theta/2) + i\sin(\theta/2))\). For our example, \(2+2i\), the trigonometric form is \(2\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))\). By taking the square root of both the modulus and the half of the argument, we obtain the two square roots, reflecting the principle that every non-zero complex number has two distinct square roots in the complex plane.
Understanding these concepts not only assists in finding the square roots of a complex number but also in laying the foundation for working with complex numbers in advanced mathematics, physics, and engineering.