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$$ z = re^{i\theta} $$

I have seen that, when specifying a complex number, most people would rather use radians as the units for $\theta.$ Is it incorrect to use degrees? Why is there a preference for radians?

ryang
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    It depends. Elementary introduction to Complex Numbers can precede Real Analysis (AKA Calculus). In Analytical Geometry, the domain of the sine and cosine functions are angles, so you don't really gain anything by measuring angles in radians rather than degrees. That is, it does not facilitate solving any Analytical Geometry problems (that I know of). For example, DeMoivre's theorem works fine against angles measured in degrees. ...see next comment – user2661923 Nov 02 '21 at 21:06
  • However, if you have already begun Real Analysis (or Calculus), then things have changed remarkably. Here, the domain of the sine and cosine functions are dimensionless real numbers, and you grapple with such concepts as the $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1.$ The term Radians is ambiguous, and can refer to the unit of measure of an angle (i.e. $\pi$ radians equals $180^\circ$) or it can refer to the dimensionless arc length of a specific arc of the unit circle. Short answer: in Calculus, it is best to stop thinking in terms of angles, measured in degrees. – user2661923 Nov 02 '21 at 21:11
  • No, it isn't incorrect to use other units (it never is fwiw), it's just that radians are the mathematically more "natural" unit. Maybe you know the formula for the arc length of a circle segment: $l = \frac{2\pi r \alpha}{360^\circ}$ where $\alpha$ is given in degrees. Note that this formula essentially has the conversion to radians built in and becomes way simpler if we express $\alpha$ in radians directly: $l = r \alpha$; so the angle itself tells you how far along the circle you've traveled. – SV-97 Nov 02 '21 at 21:13
  • This stems from the fact that radians give us a so-called "arc-length parametrization" of the circle via sine and cosine - and we like those because they have a lot of nice properties. And via eulers formula we thus also arrive at radians as the "natural unit" for the angle in the complex exponential function. – SV-97 Nov 02 '21 at 21:13
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    @user2661923 I disagree with radians being ambiguous. Even when used as an angle it's a dimensionless number (and the SI standard agrees with this - in fact all angles are technically dimensionless) – SV-97 Nov 02 '21 at 21:22
  • @SV-97 I am unsure whether I had this same disagreement with you or with someone else before. Personally, I have a hard time regarding angles as dimensionless, since the degree is a standard unit of measurement for an angle. Similarly, I regard $36$ as a dimensionless number, while $36$ inches is a measurement of length. I don't see the distinction between $36$ inches and $36$ degrees, since in both cases, you are applying a unit of measurement to a dimensionless scalar. – user2661923 Nov 02 '21 at 21:36
  • The "natural unit" depends on context. In many situations, I think rotations is a more natural unit than either degrees or radians. It depends on what properties you care about, which depends on what problem you're trying to solve... – user3716267 Nov 02 '21 at 21:38

3 Answers3

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You can use degrees… However, if you do so, then for a flat angle of $180°$ you don’t have the famous relation $$e^{i\pi}=-1$$ which holds if you use radians.

mathcounterexamples.net
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    The rebuttal here is that $e^{i\theta}$ is nothing more than syntactic sugar for $\cos(\theta) + i\sin(\theta)$. You certainly have that $\cos(180^\circ) + i\sin(180^\circ) = -1$. See the comments that I left, following the question. Please flag me with a comment if you wish to disagree. – user2661923 Nov 02 '21 at 21:16
  • @user2661923. It's not just syntactic sugar. Actually, for all $z\in\mathbb{C}$, one has $$e^z = \sum_{k=0}^{\infty} \frac{z^k}{k!}.$$ If $z=i\theta$ and $\theta$ is in radians you can use this formula directly. If $\theta$ is in degrees you have to replace $z$ with $i\pi\theta/180$ in the sum. – md2perpe Nov 02 '21 at 21:22
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    @md2perpe When I say syntactic sugar, I am assuming that that the domain of the sine and cosine functions are real numbers, rather than angles. Then $e^z = e^{x + iy} = e^x \times [\cos(y) + i\sin(y)]$, which can be shown to equal $\displaystyle \sum_{k=0}^\infty \frac{(x + iy)^k}{k!}$. From my perspective, all that the $e^z$ notation achieves is simply a convenience of notation. – user2661923 Nov 02 '21 at 21:29
  • @user2661923. And that equality is only valid if $y$ is in radians, not if it's in degrees. – md2perpe Nov 02 '21 at 21:46
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    @md2perpe No, the equality is also valid if $y$ is in neither radians nor degrees, but is instead a dimensionless real number. This is viable by my assumption that the domain of the sine and cosine functions are dimensionless real numbers. – user2661923 Nov 02 '21 at 21:48
  • @user2661923. The series $\sum_{k=0}^{\infty} \frac{\theta^{2k}}{(2k)!}$ will only converge to $\cos\theta$ if $\theta$ is given in radians (which is a dimensionless real number). So for $\theta=\pi/2$ it will converge to $0$, and for $\theta=90$ it will converge to $\approx -0.4481$. It will not converge to $\cos 90^\circ=0.$ – md2perpe Nov 02 '21 at 21:55
  • @md2perpe We are going to have to agree to disagree here. From my perspective, Radians is an ambiguous term that may refer to the unit of measurement of an angle or the dimensionless proportion of the arc length of a unit circle. Further the convergence does not depend on any use of Radians. Defining the domain of the sine and cosine functions as (dimensionless) elements in $\Bbb{R}$ works just fine. I repeat : we are going to have to agree to disagree here. – user2661923 Nov 02 '21 at 22:00
  • @user2661923. So you think that $\sum_{k=0}^{\infty} \frac{90^{2k}}{(2k)!}$ converges to $\cos 90^\circ=0$? – md2perpe Nov 02 '21 at 22:10
  • @md2perpe No, when I say that the dimensions of the sine and cosine functions are real numbers, I intend that $\sin(\pi/2) = 1$. Note that I am not saying that $\sin(\pi/2 ~\text{Radians}) = 1$, but rather that $\sin(\pi/2) = 1.$ Again, the term Radians is an ambiguous word that is used in one way by new Calculus students who still haven't grasped that the domain of the sine and cosine functions are elements of $\Bbb{R}$ and is used in another way by more experienced students of Real Analysis. – user2661923 Nov 02 '21 at 22:16
  • @user2661923. So you agree that $\theta$ in the series for $\cos\theta$ and $\sin\theta$ should be the value when the angle is measured in radians? Then you have another discussion about whether the arguments to $\cos$ and $\sin$ should be thought of as angles (which is done in trigonometry) or just numbers (which is the case in analytics), and also regarding radians as a measure of angle vs. a quotient of arc length by radius, right? – md2perpe Nov 02 '21 at 22:26
  • @md2perpe No, in Real Analysis, the concept of angles doesn't enter into the discussion of the domain of the sine and cosine functions. Further, the use of the term Radians is totally unnecessary, but does provide a reasonable crutch for the beginning Calculus student to transition into thinking that the domain of the Sine and Cosine functions are Real Numbers. In Real Analysis, it is harmless but unnecessary to use Radians to refer to the dimensionless proportion of the arc length of a unit circle, whose circumference is $(2\pi)$. ...see next comment – user2661923 Nov 02 '21 at 22:34
  • @md2perpe I am reminded of the axioms in "Calculus 2nd Edition Volume 1" (1966: Tom Apostol), that the sine and cosine functions have to satisfy. Nowhere in these axioms will you see the term Radians. – user2661923 Nov 02 '21 at 22:36
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The formula $z=re^{i\theta}$ is based on the observation that $$\begin{align} e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} &= \sum_{n=0\\n\text{ even}}^{\infty} \frac{(i\theta)^n}{n!} + \sum_{n=0\\n\text{ odd}}^{\infty} \frac{(i\theta)^n}{n!} \\ &= \sum_{k=0}^{\infty} \frac{(-1)^k\theta^{2k}}{(2k)!} + i \sum_{k=0}^{\infty} \frac{(-1)^k\theta^{2k+1}}{(2k+1)!} = \cos\theta + i\sin\theta, \end{align}$$ which is only valid if $\theta$ is in radians.

From this we get $$ z=x+iy=r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta)=re^{i\theta}. $$

Radians is the natural unit of angle in mathematics. You should learn and use them almost everywhere in mathematics.

md2perpe
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  1. The angle $73^\circ=1.274\,$rad is a dimensionless quantity: there's no dimension (in the sense of dimensional analysis) involved in measuring $\frac{73}{360}$ of a full turn/circle (a turn, like a cycle, is dimensionless), while the length dimensions get cancelled out when dividing arc length by radius. It's a misconception that unit and dimension are synonyms.

  2. The input of the natural (radian) trigonometric function $\sinθ$ is not actually an angle, but a number corresponding to some angle. E.g., $\theta$ might be $2.51,$ which corresponds to—but doesn't equal—$2.51\,\mathrm{rad}=144^\circ.$

  3. $$z=r(\cos\theta+i\sin\theta)$$ is the polar form of a complex number. By Euler's formula, $$z=re^{iθ},\tag1$$ while by the Taylor series for sine and cosine, $$z=r\left(\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\ldots\right)+i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\ldots\right)\right)$$$$=r\left(1+(i\theta)+\frac{(i\theta)^2}{2!}-\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+\ldots\right).\tag2$$

    Complex Analysis (specifically, applications involving $\exp()$ and $e^{()}$) has been developed based on the natural circular measure radian and the natural (radian) trigonometric functions. For example, the various derivations of formulae $(1)$ and $(2)$ all involve the natural trigonometric functions.

    Electing to work in ‘degrees’ is fine if we avoid $\exp()$ and $e^{()}$ and stick to purely geometric analyses/perspectives. In this case, $e^{iθ}$ is not a useful shorthand for $(\cos\theta+i\sin\theta)$ due to potential for confusion; stick to the $\mathrm{cis}\,\theta$ notation instead.

ryang
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