Problem 88
Question
Finding the \(n\) th Roots of a Complex Number \(\operatorname{In}\) Exercises \(81-96,(\) a) use the formula on page 446 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the roots in standard form. Fourth roots of $$625i$$
Step-by-Step Solution
Verified Answer
The four roots of the complex number \(625i\) are \(5\sqrt[4]{5}( \cos(22.5°) + i \sin(22.5°) )\), \(5\sqrt[4]{5}( \cos(67.5°) + i\sin(67.5°) )\), \(5\sqrt[4]{5}(\cos(112.5°) + i \sin(112.5°) )\) and \(5\sqrt[4]{5}( \cos(157.5°) + i \sin(157.5°) )\).
1Step 1: Conversion to Polar Form
The first step is to change the given complex number into polar form. The given complex number is \(625i\), which can be written as \(0 + 625i\). The magnitude \(r\) (also known as the modulus) can be computed using the formula \(r = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts of the complex number respectively. The angle \(\Theta\) (argument of the complex number or angle with respect to the positive x-axis) can be calculated using the formula \(\Theta = \arctan(b/a)\). Here, \(a = 0\) and \(b = 625\), so \(r = 625\) and \(\Theta = 90°\). Thus, the polar form is \(625(\cos(90°)+i\sin(90°))\).
2Step 2: Application of the Root Formula
The root formula for a complex number states that the \(n\)th roots \(z_k\) are given by \(z_k=r^{1/n}[\cos((\Theta + 2k\pi)/n)+i\sin((\Theta + 2k\pi)/n)]\), where \(k = 0, 1, 2, . . . , n-1\). Here the \(4\)th roots are required, so \(n=4\). The value of \(r^{1/n}\) is \((625)^{1/4} = 5\sqrt[4]{5}\). The roots are: \(z_0= 5\sqrt[4]{5}(\cos((90°+ 0*180°)/4)+i\sin((90°+ 0*180°)/4))\). \(z_1= 5\sqrt[4]{5}(\cos((90°+ 180°)/4)+i\sin((90°+ 180°)/4))\). \(z_2= 5\sqrt[4]{5}(\cos((90°+ 360°)/4)+i\sin((90°+ 360°)/4))\). \(z_3= 5\sqrt[4]{5}(\cos((90°+ 540°)/4)+i\sin((90°+ 540°)/4))\). Calculate the respective trigonometric functions to find the roots.
3Step 3: Graphical Representation
It might be tough to represent the roots graphically in the real and imaginary axis, as it is more visual. However, each root should be represented as a point on the graph corresponding to its real and imaginary parts.
4Step 4: Conversion to Standard Form
The standard form of a complex number is \(a+ bi\). Convert the \(4\)th roots, found in step 2, to the standard form by applying the trigonometric values to the cosine and sine functions. Then, write each root in the form \(a + bi\).
Key Concepts
Polar Form of Complex NumbersModulus and Argument of Complex NumbersDe Moivre's Theorem
Polar Form of Complex Numbers
Understanding the polar form of complex numbers is essential for further studies in complex analysis and is particularly crucial when it comes to finding roots of complex numbers. A complex number, which is normally expressed in the form of \(a+bi\), can alternatively be represented in polar form as \(r(\cos(\Theta) + i\sin(\Theta))\), where \(r\) is the modulus and \(\Theta\) the argument of the complex number.
For the complex number \(625i\), as shown in the step by step solution, the conversion starts with identifying the real part \(a=0\) and the imaginary part \(b=625\). The modulus \(r\) is calculated as \(\sqrt{a^2+b^2}\), giving us \(\sqrt{0^2+625^2}=625\). The argument \(\Theta\) is determined using \(\arctan(\frac{b}{a})\), which, in this case, results in \(90^\circ\). After calculating the modulus and the argument, the complex number can be written in polar form as \(625(\cos(90^\circ)+i\sin(90^\circ))\).
This representation is particularly useful for multiplying, dividing, and finding powers and roots of complex numbers, as it simplifies these operations significantly.
For the complex number \(625i\), as shown in the step by step solution, the conversion starts with identifying the real part \(a=0\) and the imaginary part \(b=625\). The modulus \(r\) is calculated as \(\sqrt{a^2+b^2}\), giving us \(\sqrt{0^2+625^2}=625\). The argument \(\Theta\) is determined using \(\arctan(\frac{b}{a})\), which, in this case, results in \(90^\circ\). After calculating the modulus and the argument, the complex number can be written in polar form as \(625(\cos(90^\circ)+i\sin(90^\circ))\).
This representation is particularly useful for multiplying, dividing, and finding powers and roots of complex numbers, as it simplifies these operations significantly.
Modulus and Argument of Complex Numbers
The modulus and argument of a complex number provide another way to express the number, which is often more useful than the standard form for certain operations. The modulus, \(r\), is simply the distance of the complex number from the origin in the complex plane. It's equivalent to the magnitude of a vector and is always a non-negative real number, found by the formula \(r = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts of the complex number, respectively.
Meanwhile, the argument, \(\Theta\), represents the counterclockwise angle measured from the positive real axis to the line segment that joins the origin to the point representing the complex number. It is calculated as \(\arctan(\frac{b}{a})\), although in cases where the complex number lies on the imaginary axis (like \(625i\)), the argument takes on standard angles like \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) for positive and negative imaginary numbers, respectively.
The process of finding the modulus and argument is the foundational step before applying more advanced operations such as finding nth roots or using De Moivre's Theorem. It's essential to be comfortable with using these calculations to explore the broader aspects of complex numbers.
Meanwhile, the argument, \(\Theta\), represents the counterclockwise angle measured from the positive real axis to the line segment that joins the origin to the point representing the complex number. It is calculated as \(\arctan(\frac{b}{a})\), although in cases where the complex number lies on the imaginary axis (like \(625i\)), the argument takes on standard angles like \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) for positive and negative imaginary numbers, respectively.
The process of finding the modulus and argument is the foundational step before applying more advanced operations such as finding nth roots or using De Moivre's Theorem. It's essential to be comfortable with using these calculations to explore the broader aspects of complex numbers.
De Moivre's Theorem
De Moivre's Theorem serves as a powerful tool in complex number calculations, particularly in calculating powers and roots. According to this theorem, for any complex number \( z = r(\cos\Theta + i\sin\Theta) \), its nth power is given by \( z^n = r^n(\cos(n\Theta) + i\sin(n\Theta)) \). This relationship becomes extremely handy as it sidesteps the need for repetitive multiplication of complex numbers in trigonometric form.
Applying it to find nth roots, we modify the theorem accordingly. If we are looking for the n-th roots of a complex number, we use the associated formula: \( z_k = r^{1/n}[\cos((\Theta + 2k\pi)/n) + i\sin((\Theta + 2k\pi)/n)] \) where \(k = 0, 1, 2, ..., n-1\). The value of \(r^{1/n}\) is simply the nth root of the modulus, and \(\Theta/n\) is the division of the argument by \(n\), adjusted by adding \(2k\pi\) to ensure that all distinct nth roots are found.
In the example of finding the fourth roots of \(625i\), De Moivre's Theorem simplifies the complex multiplication process by a large margin, leading to a more straightforward computation of the roots. Such applications make De Moivre's Theorem an indispensable tool in the toolbox of anyone working with complex numbers.
Applying it to find nth roots, we modify the theorem accordingly. If we are looking for the n-th roots of a complex number, we use the associated formula: \( z_k = r^{1/n}[\cos((\Theta + 2k\pi)/n) + i\sin((\Theta + 2k\pi)/n)] \) where \(k = 0, 1, 2, ..., n-1\). The value of \(r^{1/n}\) is simply the nth root of the modulus, and \(\Theta/n\) is the division of the argument by \(n\), adjusted by adding \(2k\pi\) to ensure that all distinct nth roots are found.
In the example of finding the fourth roots of \(625i\), De Moivre's Theorem simplifies the complex multiplication process by a large margin, leading to a more straightforward computation of the roots. Such applications make De Moivre's Theorem an indispensable tool in the toolbox of anyone working with complex numbers.
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