An introductory post to complex numbers.

This is an introductory post that will introduce you to the world of complex numbers. This is designed to teach the reader who doesn't know anything about it yet and to enhance the understanding of those who already know it. In addition, the problems will be designed to challenge you to see if you really know your stuff. As such, I've tagged the post with both the CosinesGroup tag as well as the TorqueGroup tag.

First off, what is ii? ii is an arbitrary unit that we mathematicians "made up" in order to explain the phenomenon of negative square roots. In particular, i=1i = \sqrt{-1}. Now, you might ask, why bother? This unit isn't just something that mathematicians decided to use because it would make us feel better about having negative numbers under square roots. Indeed, the world of complex numbers opens up endless new and exciting possibilities about numbers. But that's for another lesson.

A complex number, commonly noted with the variable zz, is one can be expressed in the form a+bia + bi for a,bRa, b \in \mathbb{R} (a and b are real numbers). Note that real numbers are complex numbers -- namely, when b=0b = 0. That is, RC\mathbb{R} \subset \mathbb{C} (the real numbers are a subset of the complex numbers). The complex conjugate of an imaginary number z=a+biz = a+bi, denoted by z\overline{z}, is equal to abia - bi.

In the complex plane, the xx axis corresponds to the real part of the number, while the yy axis corresponds to the imaginary part of the number. Thus, the number a+bia+bi corresponds to the point (a,b)(a, b) in the complex plane.

In addition, the absolute value of a complex number is given by its distance from the origin. Using the euclidean distance formula, this is equal to z=a2+b2|z| = \sqrt{a^2 + b^2}. Note that the absolute value of a complex number is positive and real -- after all, we're dealing with distance here.

Furthermore, every complex number can also be expressed in two other forms. The first I will introduce is trigonometric form. This is in the form r(cosθ+isinθ)r(\cos \theta + i \sin \theta), commonly expressed as rcisθr cis \theta. rr corresponds to the radius -- that is, the distance of the point (a,b)(a,b) from the origin of the complex plane. Thus, we know that r=a2+b2r = \sqrt{a^2 + b^2}. Additionally, θ\theta corresponds to the angle that is formed by the point (a,b)(a, b), the origin, and the positive x-axis. Thus, it can be calculated as tanθ=ba\tan \theta = \frac{b}{a}.

One of the most useful applications of this theorem is the fact that the following holds:

(r1cisθ1)(r2cisθ2)=r1r2cis(θ1+θ2)(r_1 cis \theta_1 )(r_2 cis \theta_2) = r_1 r_2 cis (\theta_1 + \theta_2). In addition, in the special case where we are multiplying the same complex number together, we get the result of DeMoivre's Theorem that (rcisθ)n=rncisnθ(r cis \theta)^n = r^n cis n \theta. This is one of the most commonly used theorems in math problems dealing with complex numbers. Know it and understand it inside and out. That's it for now, let's get started on some problems:

  1. A complex number z|z| satisfies z+z=2+8iz + |z| = 2 + 8i. What is z|z|?

  2. A complex number zz is equal to 9+bi9 + bi, where bb is a positive real number. Given that the imaginary parts of z2z^2 and z3z^3 are equal, find bb.

  3. It is a well known fact that cos2x=cos2xsin2x,sin2x=2sinxcosx\cos 2x = cos^2 x - sin^2 x, \sin 2x = 2 \sin x \cos x. How can we use our knowledge of complex numbers to find similar forms for cos3x\cos 3x and sin3x\sin 3x?

  4. Show that tan11+tan12+tan13=π\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \pi.

  5. Let SS denote the region in the complex plane that is made up of all points zz such that z40\large{\frac{z}{40}} and 40z\large{\frac{40}{\overline{z}}} both have real and imaginary parts between 00 and 11 inclusive. What is the area of the region SS?

Problems #2 and #5 are credit to AIMEAIME.

Image credit: Wikipedia, Mandelbrot Set

#ComplexNumbers #TorqueGroup

Note by Michael Tong
7 years, 5 months ago

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Comments

Nice article!

You can use \overline{z} for z\overline{z}. :)

Sreejato Bhattacharya - 7 years, 5 months ago

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Yes we can,

Muh. Amin Widyatama - 7 years, 5 months ago

Thanks!

Michael Tong - 7 years, 5 months ago

A good next article to read is Sotiri's note in the #TorqueGroup entitled Proofs using Complex Numbers.

Calvin Lin Staff - 7 years, 5 months ago

Hi, Michael, the angle θ\theta is not equal to tan1ba\tan^{-1} \frac{b}{a}, always, because, using that you would say that the argument of 1i-1 - i is π4\frac{\pi}{4}.

Note: tanθ\tan \theta is always equal to ba\frac{b}{a}, but, θtan1ba\theta \neq \tan^{-1} \frac{b}{a} always, because the range of tan1\tan^{-1} function is fixed, and not all angles can lie in this range.

jatin yadav - 7 years, 5 months ago

  1. This doesn't need concept of complex numbers. Here all of 1, 2 and 3 are greater than or equal to 1 and hence it comes out to be pi + 0 = pi

Christopher Johnboy - 7 years, 5 months ago

Nicely presented!!Cheers!

Eddie The Head - 7 years, 5 months ago

Does anyone want to take a stab at the problems?

Michael Tong - 7 years, 5 months ago

Hints for the problems:

  1. z|z| is always real, so we know that the imaginary part of 2+8i2 + 8i must have come exclusively from zz.

  2. What is z2z^2 and z3z^3 in expanded form? What are the imaginary parts of these numbers in expanded form?

  3. DeMoivre's formula?

  4. Where in complex analysis might we see angles in the form of tan1xtan^{-1}x being added together?

  5. Split the question into two cases, and then find the intersection of there areas.

Michael Tong - 7 years, 5 months ago

  1. z = x + iy. Then x + iy + sqrt (x^2 + y^2) = 2 + 8i Thus, y = 8, x + sqrt(x^2 + 64) = 2 or, x^2 + 64 = 4 - 4x + x^2 or, x = -15 So |z| = 17

Christopher Johnboy - 7 years, 5 months ago

  1. b = 15

Christopher Johnboy - 7 years, 5 months ago

  1. We simply have to apply De-Moivre's Theorem

Christopher Johnboy - 7 years, 5 months ago

@Michael Tong Can you add this to suitable parts of the Complex Numbers Wiki? Thanks!

Calvin Lin Staff - 6 years, 8 months ago
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