Week – 1 | Complex Numbers – 1

1.1 Introduction

A Complex Number \(z\) is represented by a combination of 2 real values – \((a,b)\) across a 2D plane. The 2 axes form the ‘Real’ and ‘Imaginary’ axes, that are used to uniquely identify a Complex Number on the Complex plane – also called the Argand Plane.  Thus an algebraic representation of \(z\) is given by

\(z = a+ib\)

where \(i=\sqrt{-1}\)

Note that we refer to \(a\) & \(b\) as the real & imaginary parts of complex number \(z\) respectively.  In the Argand Plance, the vertical axis is the imaginary axis and the horizontal axis forms the real axis. Thus \(\text{Re}(z)=a\) & \(\text{Img}(z)=b\) is the way we refer to the real and imaginary parts of a complex number.

As you can see above, the complex number A has real part 2 and imaginary part 3.

We define the ‘modulus’ of a complex number as its magnitude. In other terms, the eucleidian distance of the complex number from the origin (\(z_0=0+0.i\)) is defined as the modulus and is defined as \(|z|\). Thus if we define \(z_0\) as the origin or center point of the Argand Plane, then \(|z|\) is simply the distance between the complex numbers \(z\) & \(z_0\). We must also note that in this case, distance between 2 complex points is not dependent on the choice of origin or center of a plane.

So, we define the distance between any 2 complex numbers \(z_1\) & \(z_2\) as \(|z_1-z_2|\).

Consider the following example where \(z_1=a+ib\)  and \(z_2=c+id\)

Fig 1.1.1

Consider that we need to find the distance between the 2 complex numbers. From Pythagoras Theorem, we know that \(|z_1|=\sqrt{a^2+b^2}\) considering the right angled triangle with vertices B,C,\(z_1\). Similarly we also have \(|z_2|=\sqrt{c^2+d^2}\) considering the right angled triangle with vertices B,D,\(z_2\).

To find the distance between the points \(z_1\) and \(z_2\), we draw a line parallel to the real axis from \(z_1\). Thus

Fig 1.1.2

\(|z_1-z_2|^2=(\text{Re}(z_2-z_1))^2+(\text{Img}(z_2-z_1))^2\)

Thus we have \(|z_1-z_2|=\sqrt{(a-c)^2+(b-d)^2}\).

Note that if we replace \(z_2\) by the center of the Argand plane, we get the modulus of \(z_1\) which is

\(|z_1| = \sqrt{a^2+b^2} = \sqrt{(\text{Re}(z_1))^2+(\text{Img}(z_1))^2}\)   \( —(1)\)

Modulus, or magnitude preserves equality on either sides of an equation, so if for some complex numbers \(z_1\) & \(z_2\), if \(f(z_1)=g(z_2)\) this implies \(|f(z_1)|=|g(z_2)|\). Modulus is always a positive real value, and modulues of a real negative number is always equal to the positive magnitude of that number. Also, with algebraic simplification, we can prove \(|z_1z_2|=|z_1||z_2|\), which can further be extended to \(|\prod z_i|=\prod |z_i|\).

We have thus shown that for \(z=a+ib\), \(|z|=\sqrt{a^2+b^2}\), comparing the first and third equations of \((1)\), we can show that \(|z|=|\bar{z}|\), since \(\text{Re}(z)=\text{Re}(\bar{z})\) and \(\text{Img}(z)=-\text{Img}(\bar{z})\).

 

1.2 Conjugate of a Complex Number

We define the ‘Conjugate’ of a complex number \(z\) as \(\bar{z}\) to be the reflection of the complex number \(z\) about the Real-Axis. Thus if \(z=a+ib\) then we have \(\bar{z}=a-ib\).  Another important result is \(z\bar{z}=|z|^2\) (Try proving this)

• \(\overline{z_1 \pm z_2} =\bar{z_1} \pm \bar{z_2}\)
• \(\overline{z_1z_2}=\bar{z_1}\bar{z_2}\)
• \(\overline{(\frac{z_1}{z_2})}=\frac{\bar{z_1}}{\bar{z_2}}\)
• If \(z\bar{z}=1\), we can say \(|z\bar{z}|=|1|=1\), and using the results above, we have \(|z|^2=1\). And from that we have \(|z|=1\)

 

2.1 Basic Algebraic Properties
Considering the definitions of \(z_1\) & \(z_2\) as above

• \(z_1+z_2=(a+c)+i(b+d)\) which means \(\text{Re}(z_1+z_2)=\text{Re}(z_1)+\text{Re}(z_2)\)
• \(z_1z_2=(ac-bd)+i(ad+bc)\)
• \(\frac{z_1}{z_2}=\frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2}\)
• \(i = \sqrt{-1} \). Thus \(i^2=-1\), \(i^3=-i\) and \(i^4=1\)
• \(\text{Img}(z+\bar{z}) = 0\)

 

2.2 Argument of a Complex Number

We define the tangent of the argument of a complex number as \(\tan\theta=\frac{b}{a}\). This is basically the slope of the line joining the complex number \(z\) with the origin. Thus a complex number with 0 imaginary part has argument zero – since it lies entirely on the horizontal real axis and thus the slope of the line joining it with the origin is 0. Thus the ‘argument’ of the complex number is \(\tan^{-1}(\frac{b}{a})\)

Try Your Hand:

1. Show that \(|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)\)
2. Prove the inequality \(-|z| \le \text{Re}(z) \le |z|\). Can we replace \(\text{Re}(z)\) with \(\text{Img}(z)\) in the inequality?

2.3 Euler’s Theorem and Exponential form of complex numbers

Euler’s Theorem states that

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

 

If we consider the vectorial representation of a complex number:

and we consider \(\triangle BZ_2D\) of Fig 1.1.2,

we can say that the vector segment \(\overrightarrow{z_2} =\overrightarrow{BD}+i\overrightarrow{DZ_2} \)

Thus \(\overrightarrow{z_2} = r\cos\theta+ir\sin\theta\), where \(r\) is the length of the segment \(\overrightarrow{BZ_2}\) and \(\theta\) is \(\angle Z_2BD\) – which is what we have defined as the principal argument of the complex number.

In general, from the above theory and using Euler’s theorem, a comlex number can be represented in an exponential form in the following way:

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

where \(r\) is the magnitude of \(z\) and \(\theta\) is the principal argument. We must note that when we compare 2 complex numbers \(z_1\) & \(z_2\), we compare their real and imaginary parts. So if we compare the above definition of \(z\) with what we have defined earlier we have:

\(r\cos\theta+ir\sin\theta=a+ib\iff r\cos\theta=a\) & \(r\sin\theta=b\)

Thus \(\fbox{$\displaystyle z^n = r^ne^{n\theta} = r^n(\cos n\theta+i\sin n\theta )$}\). This is known as De Moivre’s Theorem.

Importantly, we also have

This property forms the basis for nth roots of unity, which we will discuss in our third module.

One more thing, if \(z_1=e^{i\alpha}\) & \(z_2=e^{i\theta}\), then \(z_1z_2=e^{i(\alpha + \theta )} = \cos(\theta + \alpha )+i\sin (\theta + \alpha )\). In this case we say that \(z_1\) is rotated by an angle \(\theta\).

Ex – 1:

If \(z+\frac{1}{z}=2\cos \theta\), find \(|\frac{z^{2n}-1}{z^{2n}+1}|\)

Solution:

We have \(z^2-2\cos \theta z +1=0\)

This implies \(z=\frac{2\cos \theta \pm \sqrt{4\cos ^2 \theta -4}}{2} = \cos \theta \pm i\sin \theta\)

\(k = \frac{z^{2n}-1}{z^{2n}+1} = \frac{z^n-\frac{1}{z^n}}{z^n+\frac{1}{z^n}}\)

To evaluate this, we first proceed with the positive sign of the discriminant part of \(z\).

Thus we have \(k =\frac{ (\cos \theta +i\sin \theta )^n – (\cos \theta -i\sin \theta )^n}{ (\cos \theta +i\sin \theta )^n + (\cos \theta -i\sin \theta )^n} = \frac{2i\sin n\theta}{2 \cos n\theta} = i\tan n\theta\).

Taking the negative value of the discriminant, we have \(k = \tan n\theta\).

The modulus in either case turns out to be \(\tan n\theta\).

 

Ex -2

Let \(z\) & \(w\) be 2 complex numbers that are non-zero and satisfy the following 2 conditions:

\(|z|=|w|\) 

\(arg(z)+arg(w)=\pi\). Find the value of \(z+\overline{w}\)

Solution:

If we assume \(w=|w|(\cos \theta +i\sin \theta)\) then \(z = |z|(-\cos \theta +i\sin \theta) = -|w|(\cos \theta -i\sin \theta) = -\bar{w}\). Thus the required result is \(0\).

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