Problem 1 - Complex numbers from A to...Z

Question

Find the square roots of the following complex numbers: a) \(z=1+i\) b) \(z=i\); c) \(z=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} ;\) d) \(z=-2(1+i \sqrt{3})\) e) \(z=7-24 i\)

Step-by-Step Solution

Answer

Based on the above solution, find the square roots of the following complex numbers: a) \(1 + i\) b) \(i\) c) \(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\) d) \(-2(1 + i\sqrt{3})\) e) \(7 - 24i\) Solution: a) Square roots: \(1 + i\) and \(-1 - i\) b) Square roots: \(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\) c) Square roots: \(1\) and \(-1\) d) Square roots: \(\sqrt{2}(1 + i\sqrt{3})\) and \(-\sqrt{2}(1 + i\sqrt{3})\) e) Square roots: \(3 - 4i\) and \(-3 + 4i\)

Step 1: Convert the given complex numbers into polar form

To convert a complex number from rectangular form (a + bi) into polar form (r(cosθ + isinθ)), use the formulas: \(r = \sqrt{a^2 + b^2}\) \(θ = \arctan{(\frac{b}{a})}\) We will perform this step for all the given complex numbers: a) \(z = 1 + i\) \(r = \sqrt{(1^2 + 1^2)} = \sqrt{2}\) \(θ = \arctan{(\frac{1}{1})} = \arctan{(1)} = \frac{π}{4}\) b) \(z = i\) \(r = \sqrt{(0^2 + 1^2)} = 1\) \(θ = \arctan{(\frac{1}{0})} = \frac{π}{2}\) c) \(z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\) \(r = \sqrt{(\frac{1}{2} + \frac{1}{2})} = 1\) \(θ = \arctan{(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}})} = \arctan{(1)} = \frac{π}{4}\) d) \(z = -2(1 + i\sqrt{3})\) \(r = \sqrt{(-2^2 + (-2\sqrt{3})^2)} = 4\) \(θ = \arctan{(\frac{-2\sqrt{3}}{-2})} = \arctan{(\sqrt{3})} = \frac{π}{3}\) e) \(z = 7 - 24i\) \(r = \sqrt{(7^2 + (-24)^2)} = 25\) \(θ = \arctan{(\frac{-24}{7})} ≈ -1.3258\)

Step 2: Apply De Moivre's theorem

We will now apply De Moivre's theorem to find the square roots of the given complex numbers by finding the values of \(n\) and \(k\). For each given complex number, the formula is: \(z^{1/2} = r^{\frac{1}{2}} (\cos{\frac{θ + 2kπ}{2}} + i\sin{\frac{θ + 2kπ}{2}}), k = 0, 1\) Calculate the square roots for the given complex numbers: a) \(z^{1/2} = (\sqrt{2})^{\frac{1}{2}} (\cos{\frac{π/4 + 2kπ}{2}} + i\sin{\frac{π/4 + 2kπ}{2}})\) 바로가기값(k 및 k = 경우)0 =0 및 1입니다(1 % 연결지점(k 및 k = 경우) 0 및 1 b) \(z^{1/2} = 1^{\frac{1}{2}} (\cos{\frac{π/2 + 2kπ}{2}} + i\sin{\frac{π/2 + 2kπ}{2}})\) Two values of k are: k = 0 and k = 1 c) \(z^{1/2} = 1^{\frac{1}{2}} (\cos{\frac{π/4 + 2kπ}{2}} + i\sin{\frac{π/4 + 2kπ}{2}})\) Two values of k are: k = 0 and k = 1 d) \(z^{1/2} = 4^{\frac{1}{2}} (\cos{\frac{π/3 + 2kπ}{2}} + i\sin{\frac{π/3 + 2kπ}{2}})\) Two values of k are: k = 0 and k = 1 e) \(z^{1/2} = 25^{\frac{1}{2}} (\cos{\frac{-1.3258 + 2kπ}{2}} + i\sin{\frac{-1.3258 + 2kπ}{2}})\) Two values of k are: k = 0 and k = 1

Step 3: Calculate the square roots

Now, with the values of k, calculate the square roots for the given complex numbers: a) \(k=0: z^{1/2} = 1 + i\) \(k=1: z^{1/2} = -1 - i\) b) \(k=0: z^{1/2} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\) \(k=1: z^{1/2} = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\) c) \(k=0: z^{1/2} = 1\) \(k=1: z^{1/2} = -1\) d) \(k=0: z^{1/2} = \sqrt{2}(1 + i\sqrt{3})\) \(k=1: z^{1/2} = -\sqrt{2}(1 + i\sqrt{3})\) e) \(k=0: z^{1/2} = 3 - 4i\) \(k=1: z^{1/2} = -3 + 4i\)

Key Concepts

Complex Numbers

Complex numbers form the backbone of various advanced fields in mathematics and engineering. These numbers comprise a real part and an imaginary part, represented typically as a + bi, where a is the real part and bi is the imaginary part. It's key to know that i is the imaginary unit and is defined by the property i^2 = -1.

Understanding complex numbers requires familiarity with the Cartesian plane where they can be plotted. The real part, a, corresponds to the x-axis, while the imaginary part, b, corresponds to the y-axis. This graphical representation can significantly aid in visualizing the operations involving complex numbers, such as addition, subtraction, and particularly, multiplication, and finding roots. The concept of complex conjugates is also crucial, where for a complex number a + bi, its conjugate is a - bi; this helps in simplifying the division of complex numbers.

De Moivre's Theorem

De Moivre's theorem is a powerful tool in complex analysis, named after the French mathematician Abraham de Moivre. This theorem links complex numbers with trigonometry and makes finding powers and roots of complex numbers simpler.

Stated formally, De Moivre's theorem says that for any complex number represented in trigonometric form (also known as polar form) as r(cos(θ) + i sin(θ)), and for any integer n, the n-th power of the complex number is given by (r(cos(θ) + i sin(θ)))^n = r^n (cos(nθ) + i sin(nθ)). This is critical when dealing with exercises that involve finding the roots of complex numbers, since it can be extended to fractional exponents. Therefore, to find the square roots as in the exercise, we can use this theorem with n = 1/2.

Polar Form of Complex Numbers

The polar form of a complex number is another way of representing complex numbers, very useful especially when we deal with multiplication, division, and finding powers and roots. Each complex number in polar form corresponds to a point in the polar coordinate system. It consists of a radius, r, and an angle, θ, and is expressed as r(cos(θ) + i sin(θ)), where r is the modulus of the complex number and θ is the argument.

The modulus or absolute value r is found by the formula √(a^2 + b^2), and it represents the distance from the origin to the point in the complex plane. Meanwhile, the argument θ is the angle in standard position that the line segment connecting the origin with the point makes with the positive x-axis, usually measured in radians. It can be determined using the arctangent function, often requiring attention for the correct quadrant.

Trigonometric Form of Complex Numbers

The trigonometric form of a complex number is essentially the same as its polar form; however, it emphasizes the connection with trigonometry. The trigonometric form explicitly displays the link between a complex number and the sine and cosine of an angle, written as r(cos(θ) + i sin(θ)).

The ability to rewrite complex numbers in this manner allows for an intuitive understanding of the complex plane as a series of circles, with the modulus determining the size of the circle, and the argument indicating the position on the circle. This form is especially useful with De Moivre's theorem since trigonometric properties can be used to calculate powers and roots of complex numbers. When it comes to square roots, as shown in the step-by-step solutions, this trigonometric form simplifies the process and results in a pair of complex conjugate roots for each original complex number.