Question

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ 1^{1 / 4} $$

Short answer

Answer: The fourth roots of the complex number 1 are: 1. \(1(\cos(0) + i\sin(0))\). 2. \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(1(\cos(1) + i\sin(1))\). 4. \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).

Step-by-step solution

Convert the given complex number to polar form

To find the roots, we first need to convert the given complex number to its polar form. The complex number is 1, which can be represented as \(1 + 0i\). In polar form, we represent a complex number as \(r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the magnitude and \(\theta\) is the angle. The magnitude (r) can be calculated using the formula \(r =\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts of the complex number, respectively. In this case, \(a = 1\) and \(b = 0\). $$ r = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 $$ The angle \((\theta)\) can be calculated using the formula \(\theta = \arctan(\frac{b}{a})\). In this case, since \(a = 1\) and \(b = 0\), the angle is: $$ \theta = \arctan(\frac{0}{1}) = 0 $$ So, the polar form of the given complex number is \(1(\cos(0) + i\sin(0))\).

Use De Moivre's theorem to find the fourth roots

According to De Moivre's theorem, to find the n-th root of a complex number in polar form, we should divide the angle \((\theta)\) by n and raise the magnitude \((r)\) to the power of \(\frac{1}{n}\). In this case, we want to find the fourth roots, so \(n = 4\). For each root, the angle will be: $$ \theta_k = \frac{(2k + 0)}{4} = \frac{2k}{4}, \qquad k = 0, 1, 2, 3 $$ Also, the magnitude of the roots will be: $$ r_k = (1)^{\frac{1}{4}} = 1 $$ Now we can find the fourth roots: 1. \(k = 0\): \(\theta_0 = \frac{2\cdot 0}{4} = 0\), so the root is \(1(\cos(0) + i\sin(0))\). 2. \(k = 1\): \(\theta_1 = \frac{2\cdot 1}{4} = \frac{1}{2}\), so the root is \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(k = 2\): \(\theta_2 = \frac{2\cdot 2}{4} = 1\), so the root is \(1(\cos(1) + i\sin(1))\). 4. \(k = 3\): \(\theta_3 = \frac{2\cdot 3}{4} = \frac{3}{2}\), so the root is \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).

Writing the roots in rectangular form (optional)

If you need the roots in rectangular form (a + bi), you can use the trigonometric identities: $$ \cos(\frac{n\pi}{4}) = \frac{a}{\sqrt{1 + a^2}} $$ and $$ \sin(\frac{n\pi}{4}) = \frac{a\sqrt{1 - a^2}}{1+a^2}. $$ However, since the question doesn't specifically ask for the roots in rectangular form, we'll leave them in polar form. So, the fourth roots of the given complex number 1 are: 1. \(1(\cos(0) + i\sin(0))\). 2. \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(1(\cos(1) + i\sin(1))\). 4. \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).