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I am seeing that complex numbers can be "rotated" through the complex plane but how can we show this in general?

For example let's say I start with the point $z_0 = a + bi$ in quadrant I (assuming it even makes sense to talk quadrants in the complex plane).

If I multiply it by $i$ I get $z_1 = ai + bi^2 = -b + ai$ which means going to the left $b$ positions on the real axis and then up $a$ units on the imaginary axis (quadrant II).

Multiply by $i$ again and $z_2 = -a - bi$, over to the left $a$, down $b$ (quadrant III).

Multiply by $i$ again and $z_3 = b - ai$, over to the right $b$, down $a$ (quadrant IV).

Multiply by $i$ again and $z_4 = a + bi = z_0$, so we're back to the start again.

But does this "90-degree rotational quality" work for any multiplier? What if we want to rotate by an arbitrary degree? How do we show how the two relate?

Now I know on the original unit circle we can define $\cos(\theta)$ and $\sin(\theta)$ as the ratios of sides of a right triangle, or also the $x$ and $y$ coordinates of where that triangle meets the unit circle.

Do we say it's analogous to defining such a triangle in the complex plane with $z = a + bi = \cos(\theta) + \sin(\theta)i$ as $x$ and $y$ coordinate again?

I feel like I'm almost there, trying to understand how $e^{ix}$ comes into play, but I want to understand how these concepts relate so I can understand what we're defining and how the operations work and where they come from.

user525966
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  • Are you familiar with the theory of Taylor series? Do you know the Taylor series expansions of $\cos(x)$, $\sin(x)$, and $\mathrm{e}^x$? I am afraid that without some knowledge of calculus, you are likely going to have to regard these things as mysterious definitions... – Xander Henderson Feb 21 '18 at 18:20
  • As to the question of rotation in general, how much linear algebra do you know? Complex numbers can be identified with a certain collection of $2\times 2$ matrices with real entries. Under this identification, the geometry is perhaps more clear. – Xander Henderson Feb 21 '18 at 18:23
  • @XanderHenderson I have heard of them, I've used them, but I want to understand them and how they're derived too – user525966 Feb 21 '18 at 18:45

5 Answers5

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Suppose we have:

$w = \cos \theta + i\sin\theta\\ z = \cos \phi + i\sin\phi$

And we multiply them together:

$wz = $$(\cos \theta + i\sin\theta)(\cos \phi + i\sin\phi)\\ (\cos\theta\cos\phi - \sin\theta\sin\phi) + i(\sin\theta\cos\phi + \cos\theta\sin\phi)\\ \cos(\theta+\phi) + i\sin(\theta + \phi)$

We are working with complex numbers unit magnitude. i.e. $|z| = 1$ but the concept still holds as any complex number can be written $z = |z|(\cos\theta + i\sin \theta).$ $\theta$ is called "the argument" of the complex number.

Multiplying complex numbers adds the arguments.

If an opperation of multiplication behaves like an operation of addition that is a property of exponentials.

We define $e^{i\theta} = \cos \theta + i\sin \theta$

$z = |z|e^{i\theta}\\w =|w| e^{i\phi}\\zw = |zw| e^{i(\theta+\phi)} = |zw|(\cos(\theta+\phi) + i\sin(\theta+\phi))$

Taylor series...

Background

$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}\\ \cos x = \sum_\limits{n=0}^\infty \frac {x^{2n}}{2n!}\\ \sin x = \sum_\limits{n=0}^\infty \frac {x^{2n+1}}{(2n+1)!}$

or:

$e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{6} +\cdots\\ \cos x = 1 - \frac {x^2}{2} + \cdots\\ \sin x = x - \frac {x^3}{6} + \cdots$

$e^{ix} = 1 + ix + \frac {(ix)^2}{2} + \frac {(ix)^3}{6} +\cdots\\ e^{ix} = 1 + ix - \frac {x^2}{2} - i\frac {x^3}{6} + \frac {x^4}{4!} + \cdots$

Collect the real terms and the imaginary terms...

$e^{ix} = $$(1 - \frac {x^2}{2} + \frac {x^4}{4!}-\cdots ) + (ix - i\frac {x^3}{6} + i\frac {x^5}{5!} - \cdots)\\ \cos x + i\sin x$

Doug M
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  • Was totally on board with you until "we define..." I don't understand why $e$ is suddenly brought on board, why it's defined that way, how we're using it, etc. – user525966 Feb 21 '18 at 18:49
  • Also it is a little strange to me (although I'm sure it's standard) to write something like $z = |z|(\cos(\theta) + i \sin(\theta))$. We're taking the magnitude of $z$... but inside the expression where we're defining $z$ itself. It feels a little bit like an infinite/recursive definition. Or is it because it's sort of like using the other definition $z=a+bi$ in-place, $z = |a + bi|(\cos(\theta) + i \sin(\theta))$? – user525966 Feb 21 '18 at 18:52
  • I was thinking about leaving that out of it. But you seemed to be interested in where this $e^{ix}$ comes from. And most complex analysis books waive it away, by defining it in this way. It can be derived, fairly straight forward using caclulus and Taylor series. With much difficulty using purely algebra and trig. – Doug M Feb 21 '18 at 18:53
  • $z = a + bi = |z|(\cos\theta + i\sin \theta), |z| = \sqrt {a^2 + b^2}, \theta = \arctan \frac ba$ and with a little bit of inspection you should be able to convince yourself that if $z \ne 0$ they can be put into 1 to 1 correspondence. – Doug M Feb 21 '18 at 18:56
  • I agree, that makes sense. How does the Taylor series show that this is a correct derivation with $e$? Taylor representation for $e, \cos, \sin$ and showing equivalence? Did Euler figure this out by using Taylor series on the trig functions and noticing it could be stated as $e^{ix}$? – user525966 Feb 21 '18 at 18:59
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    I have put in an update with the Taylor series derivation. As I understand it, it this was the route that Euler took to discover his identity. – Doug M Feb 21 '18 at 19:17
  • That's so cool... now I just have to understand Taylor series! Thanks – user525966 Feb 21 '18 at 19:24
  • @DougM In your comment you said that $\theta=\arctan \frac{b}{a}$, but this gives an incorrect answer for a lot of pairs $(a,b)$. For the correct function to find $\theta$, see, e.g. "atan2". – Mark S. Feb 22 '18 at 14:21
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A complex number $z$ can be written in the form (Euler's Identity)

$$ z=re^{i\theta} = r(cos(\theta)+isin(\theta)) $$

Where the rotation comes from is multiplying $z$ by $e^{i\alpha}$

$$ ze^{i\alpha} = re^{i(\theta+\alpha)} = r(cos(\theta+\alpha)+isin(\theta+\alpha)) $$

Samuel
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You can write a complex number as $z=re^{i\phi}$. If I multiply this by another number, $z_1=r_1e^{i\phi_1}$, I get $z_1z=r_1re^{i(\phi_1+\phi)}$. We can understand this as a scaling by a factor of $r_1=|z_1|$, and a rotation by $\phi_1=\arg(z_1)$.

Just to verify this, you can start from the trigonometric representation. $z=r\cos\theta+ir\sin\theta$, and $z_1=r_1\cos\theta_1+ir_1\sin\theta_1$. Then $$z_1z=r_1r(\cos\theta\cos\theta_1-\sin\theta_1\sin\theta)+ir_1r(\cos\theta\sin\theta_1+\sin\theta\cos\theta_1)=r_1r\cos(\theta+\theta_1)+ir_1r\sin(\theta+\theta_1)$$

Andrei
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Suppose $z \neq 0$ then $z = |z| {z \over |z|}$. Since ${z \over |z|}$ lies on the unit circle, there is some $\theta$ such that ${z \over |z|} = \cos \theta+ i\sin \theta$, and so ${z \over |z|} = e^{i \theta}$. Since $|z| >0$, if we let $\alpha = \log |z|$ we have $|z| = e^\alpha$.

Combining we have $z = |z| {z \over |z|} = e^\alpha e^{i \theta} = e^{\alpha + i \theta}$.

copper.hat
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There are mainly two questions here : 1) how does rotation link with multiplication of complex numbers 2) how does $e$ figure out in this story.

Answer to the first question is simple. Complex numbers are just ordered pairs of real numbers and hence can be represented as points in a plane (the usual coordinate geometry stuff, but here we prefer to call it the Argand diagram). Next you should be able to prove yourself using elementary geometric arguments that if $z=a+ib$ is represented by point $P$ and $O$ is the origin then the complex number $z(\cos\theta+i\sin\theta), \theta\in\mathbb{R} $ is represented by a point $Q$ such that the the line segment $OQ$ can be obtained by rotating $OP$ by an angle $\theta$ around origin. If you are familiar with the formulas for rotation from coordinate geometry this will seem obvious to you.

The appearance of $e$ in this context can be initially assumed mainly as a mnemonic and later when the need arises one can develop a proper theory of complex exponents. Thus the product formula $$(\cos\theta +i\sin \theta) (\cos\phi+i\sin\phi) =\cos(\theta+\phi) +i\sin(\theta+\phi) $$ can be easily remembered if one adopts a convention that the expression $\cos\theta+i\sin \theta $ can be written as $e^{i\theta} $. In fact since this is just a mnemonic we can altogether use the notation $\xi^{\theta} $ where one does not need to bother about what $\xi$ is. The important point is that the function $F(\theta) =\cos\theta+i\sin\theta$ satisfies the fundamental property $F(\theta) F(\phi) =F(\theta+\phi) $ and thus can be best represented as some sort of exponential function.

Later when we have developed the machinery of complex numbers in some detail we may well try to define the usual functions like $\sin, \cos, \log, \exp$ for complex numbers then the equation $$e^{i\theta} =\cos\theta +i\sin\theta$$ acquires a genuine meaning and is no longer just a tool for remembering multiplication of complex numbers. Often textbooks make the jump from the mnemonic device to complex exponential without the necessary theoretical development and that creates utter confusion.

I don't want to go in the details of how the theory of complex exponential function can be developed. Typical approach is to use either the Taylor series $e^z=\sum_{n=0}^{\infty}z^n/n!$ or just define $e^z=e^{x+iy} =e^x(\cos y+i\sin y) $ but my favorite one is the most elementary definition $$e^z=\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n$$

Paramanand Singh
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