Homework 5.5 Study Notes

Homework 5.5 Overview

No Due Date.
Contains problems on complex numbers, which are numbers of the form $a + bi$ , where $a$ and $b$ are real numbers, and $i$ is the imaginary unit satisfying $i^{2} = -1$ . They are fundamental in various fields of engineering, physics, and mathematics.
Topic covered during 8th and 9th weeks of the course, indicating a crucial period for understanding these concepts.
Not graded; no submission required, but problems should be reviewed meticulously for final exam preparation as similar concepts and problem types are highly likely to appear.

Problem Set on Complex Numbers

Problem 1: Finding Real and Imaginary Parts

Find the real and imaginary parts for the following complex numbers:
- (a) For $2 + 2i$ :
- Real part: $2$
- Imaginary part: $2$
- (b) For $(i + \sqrt{3})^{30}$ :
- Requires computation or application of De Moivre's Theorem. De Moivre's Theorem states that for any complex number $z = r(\cos \theta + i\sin \theta)$ (or $re^{i\theta}$ ) and any integer $n$ , $(r(\cos \theta + i\sin \theta))^{n} = r^{n}(\cos(n\theta) + i\sin(n\theta))$ . To use this, first convert the base $(i + \sqrt{3})$ into its polar form $r(\cos \theta + i\sin \theta)$ or $re^{i\theta}$ , where $r = |z| = \sqrt{(\mathrm{Real}_\mathrm{part})^2 + (\mathrm{Imaginary}_\mathrm{part})^2}$ is the magnitude and $\theta = \arg(z)$ is the argument (angle) of the complex number. After finding $r$ and $\theta$ for $(i + \sqrt{3})$ , apply De Moivre's Theorem with $n=30$ to simplify the power and extract the real and imaginary parts from the resulting expression.
- (c) For ie^{i \frac{\pi}{6}}$一脸
- Requires conversion from exponential form to rectangular form (a+bi $) to extract real and imaginary parts. Use Euler's formula, which states that$ e^{i\theta} = \cos \theta + i\sin \theta $. Therefore, expand$ e^{i \frac{\pi}{6}} $into its rectangular components using this formula. Then, perform the multiplication by$ i $(remembering that$ i^2 = -1 $) and identify the real and imaginary coefficients.</li><li>(d) For$ e^{3 - 2i} $:</li><li>Requires application of Euler's formula to find real and imaginary components. For a complex exponential of the form$ e^{a+bi} $, the property is$ e^{a+bi} = e^{a}e^{bi} $. Using Euler's formula for the$ e^{bi} $part, this becomes$ e^{a}(\cos b + i\sin b) $. Substitute$ a=3 $and$ b=-2 $to find the rectangular form$ e^3 (\cos(-2) + i\sin(-2)) $and subsequently its real and imaginary parts.</li></ul></li></ul><h6 id="1d48d825-1d7e-4d86-b890-51c6d6aff7b4" data-toc-id="1d48d825-1d7e-4d86-b890-51c6d6aff7b4" collapsed="false" seolevelmigrated="true">Problem 2: Solving Complex Equations</h6><ul><li>Find all solutions for the following equations:<ul><li>(a)$ z^{4} + 2z^{2} + 2 = 0 $:</li><li>This is a quadratic in disguise. Transform it into a standard quadratic equation by letting$ w = z^{2} $. The equation then becomes$ w^{2} + 2w + 2 = 0 $. Solve for$ w $using the quadratic formula,$ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where$ a=1, b=2, c=2 $. After finding the values for$ w $(which will be complex), determine$ z $by taking the square roots of$ w $. Remember that every non-zero complex number has two square roots.</li><li>(b)$ z^{5} = 1 + 2i $:</li><li>To find complex roots, it's easiest to work in polar form. Convert the right-hand side$ (1 + 2i) $to polar form$ r0 e^{i\phi0} $(where$ r0 = |1+2i| $and$ \phi0 = \arg(1+2i) $). Then, use the general formula for finding the$ n $-th roots of a complex number:$ zk = r0^{1/n} e^{i(\frac{\phi_0 + 2k\pi}{n})} $for$ k = 0, 1, \ldots, n-1 $. In this specific problem,$ n=5 $, so there will be five distinct solutions corresponding to$ k=0, 1, 2, 3, 4 $.</li><li>(c)$ z^{3} = i $:</li><li>Similar to (b), convert the right-hand side$ i $to polar form, which is$ 1e^{i\frac{\pi}{2}} $(since$ |i|=1 $and its argument is$ \frac{\pi}{2} $). Then apply the$ n $-th root formula mentioned in (b) with$ n=3 $. This will yield three distinct cube roots of$ i $in the complex plane.</li></ul></li></ul><h6 id="250200d1-f70a-4da5-9f1d-a01341f5b902" data-toc-id="250200d1-f70a-4da5-9f1d-a01341f5b902" collapsed="false" seolevelmigrated="true">Problem 3: Computing with a Given Complex Number</h6><ul><li>Suppose$ z = 1 + 3i $. Compute:<ul><li>(a)$ \bar{z} + |z|^{2} $:</li><li>Here,$ \bar{z} $represents the conjugate of$ z $. If$ z = a+bi $, then its conjugate$ \bar{z} = a-bi $. And$ |z| $is its magnitude or modulus, calculated as$ |z| = \sqrt{a^2+b^2} $. Thus,$ |z|^{2} = a^2+b^2 $. First, calculate the conjugate of$ z=1+3i $and the square of its magnitude. Then, sum these two results.</li><li>(b)$ Re(z + \bar{z}) $:</li><li>Add$ z $and its conjugate. For any complex number$ z = a+bi $and its conjugate$ \bar{z} = a-bi $, their sum is$ z + \bar{z} = (a+bi) + (a-bi) = 2a $. The real part of this sum,$ Re(z + \bar{z}) $, is simply$ 2a $, or$ 2Re(z) $. So, take the real part of$ z $and multiply it by 2.</li><li>(c)$ Im(e^{-z^{2}}) $:</li><li>This problem requires several steps. First, calculate$ -z^{2} $using$ z = 1+3i $. Let the result be$ -z^{2} = x + iy $. Then, convert$ e^{x+iy} $into rectangular form using the property$ e^{x+iy} = e^{x}e^{iy} $. Apply Euler's formula to$ e^{iy} $so that$ e^{x}e^{iy} = e^{x}(\cos y + i\sin y) $. Finally, extract the imaginary part of this expression, which will be$ e^{x}\sin y $.</li></ul></li></ul><h6 id="f2e43361-4655-4bc2-9133-6885ffe14bc8" data-toc-id="f2e43361-4655-4bc2-9133-6885ffe14bc8" collapsed="false" seolevelmigrated="true">Problem 4: Existence of a Complex Number</h6><ul><li>Is there a complex number$ z $that satisfies$ |z| - z = i $?<ul><li>Explanation required to explore conditions under which such an equality can hold. To solve this, let$ z = x + iy $, where$ x $and$ y $are real numbers. Substitute this into the equation:$ \sqrt{x^2+y^2} - (x+iy) = i $. Separate this complex equation into two real equations by equating the real parts and the imaginary parts on both sides:$ \sqrt{x^2+y^2} - x = 0 $(real part) and$ -y = 1 $(imaginary part). Solve this system of two equations for$ x $and$ y $to determine if such real values exist, thereby showing if a suitable complex number$ z $exists.</li></ul></li></ul><h6 id="cf141a23-44b4-4cec-91b3-c3591babe1b2" data-toc-id="cf141a23-44b4-4cec-91b3-c3591babe1b2" collapsed="false" seolevelmigrated="true">Problem 5: Finding Complex Numbers</h6><ul><li>Find all complex numbers satisfying$ z^{2} \bar{z} = z $:<ul><li>Solutions require simplifying the equation and analysis in terms of absolute value and argument. Consider two cases:<ul><li>Case 1:$ z=0 $. Substituting$ z=0 $into the equation gives$ 0^2 \cdot \bar{0} = 0 $, which simplifies to$ 0=0 $. So,$ z=0 $is a solution.</li><li>Case 2:$ z \neq 0 $. If$ z $is not zero, we can divide both sides of the equation by$ z $, leading to$ z \bar{z} = 1 $. Recall that the product of a complex number and its conjugate is equal to the square of its magnitude:$ z \bar{z} = |z|^{2} $. Therefore,$ |z|^{2} = 1 $, which implies$ |z| = 1 $(since magnitude is always non-negative). Geometrically, this means all solutions (other than$ z=0 $) lie on the unit circle in the complex plane. Alternatively, if expressed in polar form$ z = re^{i\theta} $, then$ \bar{z} = re^{-i\theta} $. Substituting these into the original equation:$ (re^{i\theta})^2 (re^{-i\theta}) = re^{i\theta} $. This simplifies to$ r^3 e^{i(2\theta - \theta)} = re^{i\theta} $, or$ r^3 e^{i\theta} = re^{i\theta} $. If$ r \neq 0 $, we can divide by$ re^{i\theta} $, resulting in$ r^2 = 1 $. Since$ r $must be positive,$ r=1 $. The argument$ \theta $can be any real number.</li></ul></li></ul></li></ul><h6 id="1b4974be-1b59-4fe1-9b79-8ac585ca4a10" data-toc-id="1b4974be-1b59-4fe1-9b79-8ac585ca4a10" collapsed="false" seolevelmigrated="true">Problem 6: Sine Function Relation</h6><ul><li>Suppose$ \theta $is such that$ \sin(\theta) \neq 0 $. Show that the sum of sines:<ul><li>$ \sin(2\theta) + \sin(4\theta) + \cdot \cdot \cdot + \sin(2(n-1)\theta) = \frac{\sin(n\theta) \sin(\theta)}{\sin((n-1)\theta)} $for$ n \ge 2 $:</li><li>This problem involves summing a series of sine functions. Complex numbers can simplify sums of trigonometric functions by relating sine functions to the imaginary part of complex exponentials. The problem guides you to use the finite geometric series property:$ 1 + r + r^{2} + \ldots + r^{N-1} = \frac{1 - r^{N}}{1 - r} $where$ r \neq 1 $. This formula describes how to sum a series where each term is multiplied by a constant ratio$ r $to get the next term.</li></ul></li></ul><h5 id="d453de78-8071-49bf-909d-36c55863428b" data-toc-id="d453de78-8071-49bf-909d-36c55863428b" collapsed="false" seolevelmigrated="true">Hints on Selected Problems</h5><ul><li>For Problem 6, consider:<ul><li>The finite geometric series with the common ratio$ r = e^{2i\theta} $. The sum of sines can be seen as the imaginary part of a geometric series of complex exponentials. Specifically, the sum can be written as$ Im(e^{2i\theta} + e^{4i\theta} + \ldots + e^{2(n-1)i\theta}) $. This is$ Im(e^{2i\theta}(1 + e^{2i\theta} + (e^{2i\theta})^{2} + \ldots + (e^{2i\theta})^{n-2})) $and then applying the geometric series sum formula. Here,$ N-1 = n-2 $, so$ N = n-1 $, and the first term is$ e^{2i\theta} $.</li><li>The property:$ e^{i\phi} - e^{-i\phi} = 2i \sin(\phi) $. This is derived directly from Euler's formula ($ e^{i\phi} = \cos \phi + i\sin \phi $and$ e^{-i\phi} = \cos \phi - i\sin \phi $). Subtracting the second from the first yields$ e^{i\phi} - e^{-i\phi} = (\cos \phi + i\sin \phi) - (\cos \phi - i\sin \phi) = 2i\sin \phi$$. This property allows for converting between complex exponentials and sine functions, which is crucial for linking the sum of complex exponentials back to the sum of sines.
- Note: You will not see this type of problem on the final, meant as additional challenge material to deepen understanding of complex number applications in series summation! It showcases the power of using complex exponentials to solve trigonometric identities.