The Magic Imaginary Numbers

Complex numbers may appear a difficult subject given the name. However, there is nothing of really complicated about complex numbers. However, they definitively add a pinch of \em magic \em in the mathematics manipulations that you can do with them!

A complex numbers are the set of number in {\mathbb{C}} defined as pair of real numbers {(a,b)} where {a} and {b}, the real part and imaginary part part of {z} are numbers in {\mathbb{R}}. The imaginary part contain the imaginary number {i} — If you will read engineering or physics books will notice that it used the letter {j} instead of {i} for the square root of {-1}, to avoid confusion with the symbol used to indicate the electric current — that was named such since it is defined to satisfy the rule  {i^2=-1} therefore it is formally written as {i=\sqrt{-1}}.

Often the following notation is used

\Re (a+ib)=a

\Im(a+ib) = b.

In this sense, we can think of a complex number as a point in the plane, the so-called Argand plane.

Screenshot 2018-11-09 at 20.18.24

Generally we just write {z=(a,b)} as {z=a+i b}, and we treat {i} as if it were an unknown. When {b} is zero, then {z=(a,0)} is just the number {a}. Two complex numbers are equal if and only if their real and imaginary parts are equal.

Mathematical operations with Complex Numbers

We make arithmetic operation with complex numbers considering the number as a polynomial in the unknown {i}, and whenever we obtain a {i^2}, we replace it by {-1}.

Addition/subtraction. We add complex numbers in the straightforward way: {(a,b)+(c,d)=(a+c,b+d)}. Given two complex numbers {=a+ib} and {=s+it}, we can define the following the arithmetic operations

(NOTE: Do not confuse in the second relation with the triangular inequality that state for the moduli: {|z+w| \leq |z| + |w|}):

\begin{aligned} z \pm w &= (a \pm s)+ i(b \pm t) \\ \overline{z+w} &= \bar{z}+ \bar{w} \end{aligned}

In the following figure, the addition {(-1+i)+(3+2i)=(3+2i)} is geometrically represented on the Argand plane.

Screenshot 2018-11-09 at 20.23.56

Multiplication. The multiplication\index{multiplication of complex numbers} of complex numbers is defined by

(a,b) \times (c,d) \overset{\text{def}}{=} (ac-bd,ad+bc)

It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further, and most importantly, {(0,1) \times (0,1) = (-1,0)}.

\begin{aligned} zw &= (as-bt)+i(at+bs) \\ |zw| &= \sqrt{(a^2+b^2) (s^2 + t^2)} = |z||w| \\ \overline{zw} &= (as-bt)-i(at-bs)={\overline z} {\overline w} \end{aligned}

The complex conjugate of {z} is defined as {\bar{z} = a - ib}}. Therefore, the Real and Imaginary part of {z} can be defined as:  {\Re{z}=\frac{1}{2}(z+\bar{z})} and {\Im{z}=\frac{1}{2i}(z-\bar{z}). }

The modulus of {z} is defined as

\displaystyle |z| = \sqrt{z\bar{z}} = \sqrt{(a + ib)(a - ib))} = \sqrt{a^2 + b^2} \ \ \ \ \ (2)

Division. Using the previous properties, we can also define the division of two complex number as follows:

\displaystyle \frac{z}{w} = \frac{z \overline w}{\overline w w} =\frac{(x+iy)(s-it)}{s^2+t^2} = \frac{(as+bt)}{s^2+t^2}+i\frac{(bs-at)}{s^2+t^2} \qquad (3)

It also follows that

\begin{aligned} \left| \frac{z}{w} \right| &= \frac{|z|}{|w|} \\ \overline {\left( \frac{z}{w} \right)} &= \frac{\overline{z}}{{\overline{w}}}\end{aligned}


Justify the following identities:

{i^3 = -i},

{i^4 = 1},

{\dfrac{1}{i} = -i},

{(3-7i)(-2-9i) = \cdots = -69-13i},

{(3-2i)(3+2i) = 3^2 - {(2i)}^2 = 3^2 + 2^2 = 13},

{\frac{1}{3-2i} = \frac{1}{3-2i} \frac{3+2i}{3+2i} = \frac{3+2i}{13} = \frac{3}{13}+\frac{2}{13}i}.

Complex Numbers in Exponential Form

We also define the exponential {e^{a+ib}} of a complex number. We do this by writing down the Taylor series and plugging in the complex number. Because most properties of the exponential can be proved by looking at the Taylor series, these properties still hold for the complex exponential. For example the very important property:

{e^{x+y} = e^x e^y}.

This means that

{e^{a+ib} = e^a e^{ib}}.

Hence if we can compute {e^{ib}}, we can compute {e^{a+ib}}. For {e^{ib}} we use the so-called Euler’s formula.

Theorem 2 (Euler’s formula)

e^{i\theta}=\cos (\theta)+i\sin (\theta) \ \ \ \ \ (4)

since {\cos(-\theta) =\cos(\theta)} and {\sin(-\theta)=-\sin(\theta)}

e^{-i\theta}=\cos(\theta)-i\sin(\theta) \ \ \ \ \ (5)

It follows that

{\displaystyle e^{a+ib} = e^a \bigl( \cos(b) + i \sin(b) \bigr) = e^a \cos(b) + i e^a \sin(b). \ \ \ \ \ (6)}

by adding and subtracting Eq. 4 and 5, we obtain the two useful formulas:

\begin{aligned} \cos{\theta} &= \frac{e^{i \theta}+e^{-i\theta}}{2} \\ \sin{\theta} &= \frac{e^{i\theta}-e^{-i\theta}}{2i} \end{aligned}

Also not that from Eq. 4, for {\theta=\pi}, we also obtain the amazing Euler’s formula


that relates five fundamental numbers of mathematics.

Exercise 2

Using Euler’s formula, check the identities:

\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text{and} \qquad \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} .

The trigonometric relations can be easily derived from the polar representation of a complex number.

Exercise 3

Using the relation:

{e^{i(x+y)}=e^{ix}e^{iy}} and {e^{i\theta}=\cos(\theta)+i\sin(\theta)}, you can derive the expressions for

{\sin(x+y)} and {\cos(x+y)}}.

Theorem 3 (De Moivre’s Theorem)

For every real number {\theta} and every positive integer {n}, we have

\displaystyle (\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta. \ \ \ \ \ (7)

Proof: We prove this theorem by induction, i.e. first we prove it for {n=1} and then we prove that if Eq. 7 holds for a particular value of {n}, then it holds for {n+1} as well. This suffices to prove the theorem for every positive integer {n.} The case {n=1} is trivial. Assume Eq. 7 holds for {n}. Then we have

\begin{aligned} (\cos \theta + i \sin \theta)^{n+1} &= (\cos \theta + i \sin \theta)^n (\cos \theta + i \sin \theta) \\ &= (\cos n\theta + i \sin n\theta) (\cos \theta + i \sin \theta), \end{aligned}\ \ \ \ \ (8)

where in the last step we used the induction hypothesis, i.e. the assumption that Eq. 7 holds for {n.} Computing the product on the right yields

\begin{aligned}  (\cos \theta + i \sin \theta)^{n+1} &= (\cos n\theta \cos \theta - \sin n\theta \sin \theta) + i (\sin n\theta \cos \theta + \cos n\theta \sin \theta) \\ &= \cos (n+1)\theta + i \sin (n+1)\theta, \end{aligned} \ \ \ \ \ (9)

where we used the addition formulas for sine and cosine in the last step. Thus, Eq. 7 holds for {n+1} as well, whence it holds for every positive integer {n}.


Exercise 4

Derive from {e^{i \theta n}=(e^{i \theta})^n} the expression of the De Moivres Theorem and use it to derive the following trigonometric expressions:


On the Argand plane, a complex numbers in polar coordinates is described using the distance {r} and the angle {\theta} as follows:

\displaystyle {z=re^{i\theta}} \ \ \ \ \ (10)

The Argand plane: Polar representation of complex numbers.

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