# 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

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

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