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Magazine Chapter 8: Problem 13
Find the two square roots of \(1+\sqrt{3} i\).
Short Answer
Expert verified
The square roots of \(1 + \sqrt{3}i\) are \(\frac{\sqrt{6}}{2} + \frac{i\sqrt{2}}{2}\) and \(-\frac{\sqrt{6}}{2} - \frac{i\sqrt{2}}{2}\).
Step by step solution
01
Express the Complex Number in Polar Form
The given complex number is \(z = 1 + \sqrt{3}i\). To convert it to polar form, we need to calculate its magnitude and angle. The magnitude \(r\) is given by \(r = \sqrt{a^2 + b^2}\), where \(a = 1\) and \(b = \sqrt{3}\). Therefore, \(r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2\).The angle \(\theta\) is given by \(\tan^{-1}\left(\frac{b}{a}\right)\). So, \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3})\). This means \(\theta = \frac{\pi}{3}\) (since \(\tan(\frac{\pi}{3}) = \sqrt{3}\)).Thus, the polar form of \(z\) is \(2 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3}\right)\).
02
Use De Moivre's Theorem for Square Roots
De Moivre's Theorem states that for a complex number \(r(\cos(\theta) + i\sin(\theta))\) and an integer \(n\), its \(n\)th roots are given by \[ r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right) \text{ for } k = 0, 1, \ldots, n-1. \]For square roots, set \(n = 2\) and \(r = 2\), calculating \(r^{1/2} = \sqrt{2}\).
03
Calculate Both Square Roots
Now we find the two square roots:- For \(k = 0\):\[ \sqrt{2} \left( \cos\left(\frac{\frac{\pi}{3} + 2(0)\pi}{2}\right) + i\sin\left(\frac{\frac{\pi}{3} + 2(0)\pi}{2}\right) \right) = \sqrt{2} \left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right). \]This simplifies to \(\sqrt{2} \left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \frac{\sqrt{6}}{2} + \frac{i\sqrt{2}}{2}\).- For \(k = 1\):\[ \sqrt{2} \left( \cos\left(\frac{\frac{\pi}{3} + 2(1)\pi}{2}\right) + i\sin\left(\frac{\frac{\pi}{3} + 2(1)\pi}{2}\right) \right) = \sqrt{2} \left( \cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6} \right). \]This simplifies to \(\sqrt{2} \left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = -\frac{\sqrt{6}}{2} - \frac{i\sqrt{2}}{2}\).
04
Rewrite the Results Clearly
Therefore, the square roots of \(1 + \sqrt{3}i\) are:1. \(\frac{\sqrt{6}}{2} + \frac{i\sqrt{2}}{2}\) 2. \(-\frac{\sqrt{6}}{2} - \frac{i\sqrt{2}}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
Complex numbers can be expressed in a form that highlights both their magnitude and direction, known as the polar form. For any complex number, which can be written as \( z = a + bi \), the polar form is given by \( z = r(\cos \theta + i \sin \theta) \). To convert a complex number to its polar form, follow these simple steps:
- Calculate the Magnitude: The magnitude \( r \) is the distance from the origin to the point \((a, b)\) on the complex plane. It is computed as \( r = \sqrt{a^2 + b^2} \).
- Determine the Angle: The angle \( \theta \) is measured from the positive x-axis to the line segment joining the origin and the point \((a, b)\). Calculate it using \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
De Moivre's Theorem
De Moivre's Theorem is a powerful tool in the world of complex numbers, particularly useful when finding powers and roots of complex numbers. According to the theorem, if you have a complex number in polar form \( r(\cos \theta + i \sin \theta) \), De Moivre鈥檚 Theorem states that its \( n \)th root can be found using:
- \( r^{1/n} (\cos \frac{\theta + 2k\pi}{n} + i\sin \frac{\theta + 2k\pi}{n}) \)
- This equation needs to be applied for each integer \( k \) from 0 to \( n-1 \) to find all \( n \)th roots.
- For \( k = 0 \), we find one root by simplifying \( \cos \frac{\pi}{6} + i\sin \frac{\pi}{6} \).
- For \( k = 1 \), we find the second root using \( \cos \frac{7\pi}{6} + i\sin \frac{7\pi}{6} \).
Square Roots
Finding the square roots of complex numbers might initially seem tricky, but using their polar form simplifies the task greatly. When a complex number \( z = a + bi \) is transformed into its polar representation, finding its square root involves:
- Applying De Moivre鈥檚 Theorem: As outlined, plug the values into the theorem for \( n = 2 \).
- Finding Two Roots: For a square root, there are always two solutions, corresponding to \( k = 0 \) and \( k = 1 \) in the theorem.
- The first root simplifies to \( \frac{\sqrt{6}}{2} + \frac{i\sqrt{2}}{2} \).
- The second root simplifies to \( -\frac{\sqrt{6}}{2} - \frac{i\sqrt{2}}{2} \).
