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Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ (1-i)^{1 / 2} $$

Short Answer

Expert verified
Answer: The square roots of the complex number (1 - i) are: $$ \left(\sqrt[4]{2} \cos \left(-\frac{\pi}{8}\right) + \sqrt[4]{2} \sin \left(-\frac{\pi}{8}\right)i\right) $$ and $$ \left(\sqrt[4]{2} \cos \left(\frac{7\pi}{8}\right) + \sqrt[4]{2} \sin \left(\frac{7\pi}{8}\right)i\right) $$

Step by step solution

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01

Convert the complex number to polar form

Convert the complex number \((1-i)\) to its polar form \((r,\theta)\): 1. Find the modulus \(r\): \(r = \sqrt{a^2 + b^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}\) 2. Find the argument \(\theta\): \(\theta = \arctan \left( \frac{b}{a} \right) = \arctan \left( \frac{-1}{1} \right)\). Since both the real and imaginary parts are positive, we are in the fourth quadrant, so \(\theta = -\frac{\pi}{4}\). Now, the complex number in polar form can be expressed as \((\sqrt{2}, -\frac{\pi}{4})\).
02

Apply De Moivre's theorem

To find the square roots, we'll apply De Moivre's theorem: \(z_{k}^{n} = (r,\theta_k)\), where \(n\) is the power of the root, and \(\theta_k\) is given by \(\frac{\theta + 2k\pi}{n}\) for \(k = 0,1,...,n-1\). In this case, \(n=2\).
03

Calculate the roots

Find the square roots for \(k = 0\) and \(k = 1\): 1. For \(k = 0\): \(\theta_0 = \frac{-\frac{\pi}{4} + 2(0)\pi}{2} = -\frac{\pi}{8}\) \(r_0 = \sqrt[4]{2}\) Root \(1: (r_0,\theta_0) =\left(\sqrt[4]{2}, -\frac{\pi}{8} \right)\) 2. For \(k = 1\): \(\theta_1 = \frac{-\frac{\pi}{4} + 2(1)\pi}{2} = \frac{7\pi}{8}\) \(r_1 = \sqrt[4]{2}\) Root \(2: (r_1,\theta_1) = \left(\sqrt[4]{2}, \frac{7\pi}{8} \right)\)
04

Convert roots back to rectangular form

Convert the roots back to the rectangular form: 1. Root \(1\): \(x_1 = r_0 \cos \theta_0 = \left(\sqrt[4]{2}\right) \cos \left(-\frac{\pi}{8}\right)\) \(y_1 = r_0 \sin \theta_0 = \left(\sqrt[4]{2}\right) \sin \left(-\frac{\pi}{8}\right)\) Therefore, first root: \(x_1 + yi_1 = \left(\sqrt[4]{2}\right) \cos \left(-\frac{\pi}{8}\right) + \left(\sqrt[4]{2}\right) \sin \left(-\frac{\pi}{8}\right)i\). 2. Root \(2\): \(x_2 = r_1 \cos \theta_1 = \left(\sqrt[4]{2}\right) \cos \left(\frac{7\pi}{8}\right)\) \(y_2 = r_1 \sin \theta_1 = \left(\sqrt[4]{2}\right) \sin \left(\frac{7\pi}{8}\right)\) Therefore, second root: \(x_2 + yi_2 = \left(\sqrt[4]{2}\right) \cos \left(\frac{7\pi}{8}\right) + \left(\sqrt[4]{2}\right) \sin \left(\frac{7\pi}{8}\right)i\). Hence, the square roots of the given complex number \((1-i)\) are: $$ \left(\sqrt[4]{2} \cos \left(-\frac{\pi}{8}\right) + \sqrt[4]{2} \sin \left(-\frac{\pi}{8}\right)i\right) $$ and $$ \left(\sqrt[4]{2} \cos \left(\frac{7\pi}{8}\right) + \sqrt[4]{2} \sin \left(\frac{7\pi}{8}\right)i\right) $$

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Form
In mathematics, polar form is an alternative way to express complex numbers. Instead of representing a complex number as \( a + bi \) in rectangular form, we express it as \( (r, \theta) \) in polar form. Here, \( r \) is the modulus and \( \theta \) is the argument. This representation emphasizes the number's magnitude and the angle it makes with the positive real axis. The conversion steps include:
  • Finding the modulus \( r \): This is calculated using the formula \( r = \sqrt{a^2 + b^2} \), where \( a \) is the real part and \( b \) is the imaginary part.
  • Finding the argument \( \theta \): Angle can be found using \( \theta = \arctan(\frac{b}{a}) \). It's essential to consider the quadrant to get the correct angle.
This format is especially useful for multiplying and dividing complex numbers due to its simplicity in handling angles.
De Moivre's Theorem
De Moivre's Theorem provides a straightforward method for raising complex numbers to a power or extracting roots. When a complex number is in polar form \( (r, \theta) \), the theorem states:\[ z^n = r^n \left( \cos(n\theta) + i\sin(n\theta) \right) \]This formula helps simplify calculations, especially when dealing with roots or powers. For roots, you calculate each potential "angle" by adjusting \( \theta \):
  • The angle for each root is computed using \( \theta_k = \frac{\theta + 2k\pi}{n} \), where \( n \) is the root power and \( k \) is each integer from \( 0 \) to \( n-1 \).
  • This provides multiple roots, neatly accounted for by changing \( k \).
De Moivre’s Theorem is a cornerstone in complex number theory, facilitating easier computation when converting from polar to rectangular forms or finding nth roots.
Rectangular Form
In the rectangular form, a complex number is expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. This form is highly intuitive and mirrors the Cartesian coordinate system.
  • Conversion to Polar: Transitioning from rectangular to polar involves deriving the modulus and argument, akin to transforming Cartesian coordinates to polar coordinates.
  • Use with Graphs: This form is useful for plotting complex numbers on the complex plane. The number \( a + bi \) represents the point \( (a, b) \).
  • Ease in Addition/Subtraction: The main advantage is its simplicity when adding or subtracting complex numbers by dealing with components separately.
While it's beneficial for these operations, the rectangular form becomes cumbersome for multiplication, division, or taking powers, where polar form excels.
Square Roots of Complex Numbers
Finding the square roots of a complex number can be challenging but is made easier using polar form and De Moivre’s Theorem. The process involves:
  • Conversion: Initially, transform the complex number to polar form \( (r, \theta) \).
  • Application of De Moivre’s Theorem: Use the theorem to find roots, particularly by adjusting \( \theta \) for each potential root situation, resulting in \( n \) unique roots for an \( n^{th} \) root scenario.
  • Rectangular Verification: After getting roots in polar form, revert to rectangular form for practical interpretation or plotting.
This method ensures all possible roots are uncovered and precisely calculated, offering insight into the nature of complex equations. Understanding these basics allows the exploration of more advanced topics in complex analysis.

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