Trigonometric Form of Complex Numbers

Except for $0,$ any complex number can be represented in the trigonometric form or in polar coordinates:

$z = r(\cos \alpha + i\cdot \sin \alpha ),$

where $\alpha \in\mbox{Arg}(z).$ $r,$ the modulus, or the absolute value of $z,$ is easy to find:

$|x + iy| = \sqrt{x^{2} + y^{2}}.$

But how do we find $\alpha ?$ As we know, $\alpha$ is not unique, but is found modulo $2\pi .$ The main value, $\mbox{arg}(z)$ belongs to the interval $[0, 2\pi ).$ Assume, $z = x + yi.$ Then \alpha is the angle formed with the $x$-axis by the radius vector of the point $(x, y)$ or the point where the latter intersects the unit circle, viz.,

$z/|z| = \cos \alpha + i\cdot \sin \alpha .$

Apparently, $\alpha$ can be found from

(1)

$\tan\alpha = y/x.$

While this is correct to a great extent, caution should be exercised. Obviously, (1) does not work for $x = 0.$ $x = 0$ is clearly an exceptional case, but even in that case we face two alternatives of assigning to $\mbox{arg}(z)$ either $\pi /2$ or $3\pi /2.$ The choice depends on the sign of $y.$ (For $y = 0,$ $z = 0$ and, as we know, $0$ is the only complex number not associated with any argument.)

$\begin{cases} \mbox{arg}(z) = \pi /2, & y \gt 0,\\ \mbox{arg}(z) = 3\pi /2, & y \lt 0. \end{cases}$

Similar consideration must be brought to bear in the general case (1), where $x \ne 0.$ Since $\tan()$ is periodic with period $\pi$ and the base interval $(-\pi /2, \pi /2),$ we'll have to account for the signs of both $x$ and $y.$

In the quadrants I and III, $\tan(t)$ is positive. In the quadrants II and IV, $tan(t) \lt 0.$ In quadrant I, $\mbox{arg}(z) = \arctan(y/x).$ But, for quadrant III, the positive $\arctan(y/x)$ must be upgraded with the addition of $\pi .$ Similarly, for $z$ in quadrant II, $\arctan(y/x)$ is negative. To bring it to the second quadrant we have to add $\pi .$ For $z$ in the fourth quadrant, $\arctan(y/x)$ is again negative, but its location is correct. To keep it there we now have to add $2\pi ,$ so that the value falls into the interval $[0, 2\pi ).$ To sum up, denote $\alpha _{0} = \tan(y/x).$ Then

$\begin{cases} \mbox{arg}(z) = \alpha _{0}, & x \gt 0, y \gt 0,\\ \mbox{arg}(z) = \alpha _{0} + \pi ,& x \lt 0, y \gt 0,\\ \mbox{arg}(z) = \alpha _{0} + \pi ,& x \lt 0, y \lt 0,\\ \mbox{arg}(z) = \alpha _{0} + 2\pi ,& x \gt 0, y \lt 0, \end{cases}$

or in a shorter form,

$\begin{cases} \mbox{arg}(z) = \alpha _{0}, & x \gt 0, y \gt 0,\\ \mbox{arg}(z) = \alpha _{0} + 2\pi , & x \gt 0, y \lt 0,\\ \mbox{arg}(z) = \alpha _{0} + \pi , & x \lt 0. \end{cases}$

The trigonometric form is intimately related to the operation of multiplication. Let $z = r(\cos \alpha + i\cdot \sin \alpha )$ and $w = s(\cos \beta + i\cdot \sin \beta ).$ Then

(2)

$\begin{align} zw&= rs(\cos \alpha + i\cdot \sin \alpha )(\cos \beta + i\cdot \sin \beta )\\ &= rs(\cos (\alpha + \beta ) + i\cdot \sin (\alpha + \beta ). \end{align}$

The latter follows from the definition of the multiplication of complex numbers and the addition formulas for sine and cosine.

$ \cos (\alpha + \beta ) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \space\mbox{and}\\ \sin (\alpha + \beta ) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta . $

(2) tells us that the modulus of the product of two numbers is the product of their moduli, something we already knew. But it also tells us something we did not: the argument of the product is the sum of the arguments of the multiplicands. (The latter of course needs to be taken modulo $2\pi.)$

(3+)

$|zw| = |z||w| \space\mbox{and}\\ \mbox{arg}(zw) = \mbox{arg}(z) + \mbox{arg}(w) \space(\mbox{mod} 2\pi ).$

Colloquially, to multiply two complex numbers one has to multiply their absolute values and add their arguments.

It is easy to see that $\mbox{arg}(z') = -\mbox{arg}(z)$ $(\mbox{mod}\space 2\pi ).$ From which we conclude that, for $w \ne 0,$

(3-)

$|z/w| = |z|/|w| \space\mbox{and}\\ \mbox{arg}(z/w) = \mbox{arg}(z) - \mbox{arg}(w)\space(\mbox{mod} 2\pi ).$

By induction, (3+) and (3-) imply de Moivre's identity: for any integer $n \ne 0,$

$(\cos \alpha + i\cdot \sin \alpha )^{n} = \cos (n\alpha ) + i\cdot \sin (n\alpha ).$

For $n = 0$ and $z \ne 0,$ we define $z^{0} = 1$ and then de Moivre's formula holds for all integer $n,$ with no exception. In general, (2) leads to Euler's formula

$\displaystyle e^{i\alpha}=\cos\alpha+i\sin\alpha,$

such that, quite naturally,

$\displaystyle e^{i\alpha}\cdot e^{i\beta} = e^{i(\alpha +\beta)}. $

In particular,

(4)

$[r(\cos \alpha + i\cdot \sin \alpha )]^{n} = r^{n}e^{in\alpha}=r^{n}[\cos (n\alpha ) + i\cdot \sin (n\alpha )].$

(As a matter of fact, see [Kline, p. 408], de Moivre has stated the eponymous formula only implicitly. Its standard form is due to Euler and was generalized by him to any real $n.)$

Next, let's think of what it takes to evaluate roots of complex numbers. It is clear form (4), that, for a given $w,$ choosing $|z| = |w|^{1/n}$ and $\mbox{arg}(z) = \mbox{arg}(w)/n$ will solve the problem, $w = z^{n}$:

$\begin{align} z^{n}&= [|w|^{1/n}(\cos (\mbox{arg}(w)/n) + i\cdot \sin (\mbox{arg}(w)/n))]^{n}\\ &= |w|^{n\cdot 1/n}(\cos (n\cdot \mbox{arg}(w)/n) + i\cdot \sin (n\cdot \mbox{arg}(w)/n))\\ &= |w|(\cos (\mbox{arg}(w)) + i\cdot \sin (\mbox{arg}(w)))\\ &= w. \end{align}$

The important thing to note is that, save for the trivial case $n = 1,$ we can get additional roots. For example, we might have chosen $\mbox{arg}(z) = (\mbox{arg}(w) + 2\pi )/n$ with exactly the same effect. Indeed, since both cosine and sine have period $2\pi ,$ we would have

$\begin{align} z^{n} &= [|w|^{1/n}(\cos ((\mbox{arg}(w) + 2\pi )/n) + i\cdot \sin ((\mbox{arg}(w) + 2\pi )/n))]^{n}\\ &= |w|(\cos (\mbox{arg}(w) + 2\pi ) + i\cdot \sin (\mbox{arg}(w) + 2\pi ))\\ &= w.\\ \end{align}$

Even more generally, for an integer k and $\mbox{arg}(z) = (\mbox{arg}(w) + 2k\pi )/n,$ we have

(5)

$\begin{align} z^{n} &= [|w|^{1/n}(\cos ((\mbox{arg}(w) + 2k\pi )/n) + i\cdot \sin ((\mbox{arg}(w) + 2k\pi )/n))]^{n}\\ &= |w|(\cos (\mbox{arg}(w) + 2k\pi ) + i\cdot \sin (\mbox{arg}(w) + 2k\pi ))\\ &= w. \end{align}$

Of course, some k will produce the same complex numbers. For example, if $m = k + n,$ then

$\begin{align} (\mbox{arg}(w) + 2m\pi )/n &= (\mbox{arg}(w) + 2(k + n)\pi )/n\\ &= (\mbox{arg}(w) + 2k\pi + 2n\pi )/n\\ &= (\mbox{arg}(w) + 2k\pi )/n + 2\pi \\ &= (\mbox{arg}(w) + 2k\pi )/n (mod 2\pi ),\\ \end{align}$

so that k and m define exactly the same complex number. Well, we see that in (5) not all numbers are different. But how many are? The answer is $n!$ Which means that, for any complex numbers $w$, the equation $z^{n} = w$ has $n$ different roots! This is all because the argument of a complex number is only defined modulo $2\pi .$ For example, if $n = 4,$ then in the sequence

$\mbox{arg}(w)/n,\\ (\mbox{arg}(w) + 2\pi )/4,\\ (\mbox{arg}(w) + 4\pi )/4,\\ (\mbox{arg}(w) + 6\pi )/4,\\ (\mbox{arg}(w) + 8\pi )/4, \ldots $

the fifth term (i.e. the one with the coefficient 8) equals modulo $2\pi $ the first term and, starting with that term, the sequence repeats itself having modulo $2\pi $ only four distinct terms.

This is not so surprising. After all, we were solving $z^{n} = w$ which is a polynomial equation of order $n.$ According to the Fundamental Theorem of Algebra, any polynomial equation with real (in fact, complex) coefficients, has exactly n roots.

We may also draw on our real numbers experience. For any positive real number a, there are two square roots, $\sqrt{a}$ and $-\sqrt{a}.$ As complex numbers, the first (i.e. positive) one has argument $0,$ the second (negative) has argument $\pi .$ I.e., for a positive real number $a,$

$\mbox{arg}(\sqrt{a} = 0,\\ \mbox{arg}(-\sqrt{a}) = \pi .$

Example

Let's find all the third roots of $w = -8 + 8i.$ In the trigonometric form, $w = 8\sqrt{2}(\cos (3\pi /4) + i\cdot \sin (3\pi /4)).$ Thus we get three distinct roots:

$2\cdot 2^{1/6}(\cos (\pi /4 + i\cdot \sin (\pi /4)),\\ 2\cdot 2^{1/6}(\cos (\pi /4 + 2\pi /3) + i\cdot \sin (\pi /4 + 2\pi /3)),\\ 2\cdot 2^{1/6}(\cos (\pi /4 + 4\pi /3) + i\cdot \sin (\pi /4 + 4\pi /3)). $

Recollecting that $\pi = 180^{\circ},$ we get the three roots in a little different form:

$2\cdot 2^{1/6}(\cos (45^{\circ}) + i\cdot \sin (45^{\circ})),\\ 2\cdot 2^{1/6}(\cos (165^{\circ}) + i\cdot \sin (165^{\circ})),\\ 2\cdot 2^{1/6}(\cos (285^{\circ}) + i\cdot \sin (285^{\circ})),$

all evenly spaced around the origin.

References

  1. T. Andreescu, D. Andrica, Complex Numbers From A to ... Z, Birkhäuser, 2006
  2. C. W. Dodge, Euclidean Geometry and Transformations, Dover, 2004 (reprint of 1972 edition)
  3. Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994
  4. M. Kline, Mathematical Thought From Ancient to Modern Times, v. 2, Oxford University Press, 1972
  5. E. Landau, Foundations of Analisys, Chelsea Publ, 3rd edition, 1966

Complex Numbers

  1. Algebraic Structure of Complex Numbers
  2. Division of Complex Numbers
  3. Useful Identities Among Complex Numbers
  4. Useful Inequalities Among Complex Numbers
  5. Trigonometric Form of Complex Numbers
  6. Real and Complex Products of Complex Numbers
  7. Complex Numbers and Geometry
  8. Plane Isometries As Complex Functions
  9. Remarks on the History of Complex Numbers
  10. Complex Numbers: an Interactive Gizmo
  11. Cartesian Coordinate System
  12. Fundamental Theorem of Algebra
  13. Complex Number To a Complex Power May Be Real
  14. One can't compare two complex numbers
  15. Riemann Sphere and Möbius Transformation
  16. Problems

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