Amplitude of a Complex Number

Here you will learn how to find argument or amplitude of a complex number with examples.

Let’s begin – 

Amplitude of a Complex Number (Argument of Complex Number)

Let z = x + iy, Then,

The angle \(\theta\) which OP makes with the positive direction of x-axis in anticlockwise sense is called the argument or amplitude of complex number z.amplitude

It is denoted by arg(z) or amp(z).

From Figure, we have

\(tan\theta\) = \(PM\over OM\) = \(y\over x\) = \(Im (z)\over Re (z)\)

\(\implies\) \(\theta\) = \(tan^{-1}({y\over x}\))

This angle \(\theta\) has infinitely many values differing by multiples of \(2\pi\). The unique value of \(\theta\) such that -\(\pi\) < \(\theta\) \(\le\) \(\pi\) is called the principal value of the amplitude or principal argument.

In order to find the argument of a complex number z = x + iy for different quadrants, we use the following algorithm.

Algorithm :

1). Find the value of \(tan^{-1}| y/x |\) lying between 0 and \(\pi/2\). Let it be \(\alpha\).

2). Determine in which quadrant the point P(x, y) belongs.

(i) If P(x, y ) belongs to the first quadrant, then arg(z) = \(\alpha\).

(ii) If P(x, y ) belongs to the second quadrant, then arg(z) = \(\pi – \alpha\).

(iii) If P(x, y ) belongs to the third quadrant, then arg(z) = -(\(\pi – \alpha\)) or \(\pi + \alpha\).

(iv) If P(x, y ) belongs to the fourth quadrant, then arg(z) = -\(\alpha\) or \(2\pi – \alpha\).

Example : Find the argument of the following complex numbers.

(i) \(1 + i\sqrt{3}\)

(ii) \(-2 + 2i\sqrt{3}\)

(iii) \(-\sqrt{3} – i\)

(iv) \(2\sqrt{3} – 2i\)

Solution

(i) Let  z = \(1 + i\sqrt{3}\) and let \(\alpha\) be the acute angle given by

\(\alpha\) = \(tan^{-1}|{Im(z)\over Re(z)}|\) = \(tan^{-1}|{\sqrt{3}\over 1}|\)

\(\implies\) \(\alpha\) = \(\pi\over 3\)

Since, Re(z) > 0 and Im(z) > 0. So, it lies in first quadrant 

\(\therefore\)  arg(z) = \(\alpha\) = \(\pi\over 3\)

(ii) Let  z = \(-2 + 2i\sqrt{3}\) and let \(\alpha\) be the acute angle given by

\(\alpha\) = \(tan^{-1}|{Im(z)\over Re(z)}|\) = \(tan^{-1}|{2\sqrt{3}\over -2}|\)

\(\implies\) arg(z) = \(\alpha\) = \(\pi\over 3\)

Since, Re(z) < 0 and Im(z) > 0. So, it lies in second quadrant 

\(\therefore\) arg(z) = \(\pi – \alpha\) = \(\pi\) – \(\pi\over 3\) = \(2\pi\over 3\)

(iii) Let  z = \(-\sqrt{3} – i\) and let \(\alpha\) be the acute angle given by

\(\alpha\) = \(tan^{-1}|{Im(z)\over Re(z)}|\) = \(tan^{-1}|{-1\over -\sqrt{3}}|\)

\(\implies\) \(\alpha\) = \(\pi\over 6\)

Since, Re(z) < 0 and Im(z) < 0. So, it lies in third quadrant 

\(\therefore\)  arg(z) = -(\(\pi – \alpha\)) = -(\(\pi\) – \(\pi\over 6\)) = \(-5\pi\over 6\)

(iv) Let  z = \(2\sqrt{3} – 2i\) and let \(\alpha\) be the acute angle given by

\(\alpha\) = \(tan^{-1}|{Im(z)\over Re(z)}|\) = \(tan^{-1}|{-2\over 2\sqrt{3}}|\)

\(\implies\) \(\alpha\) = \(\pi\over 6\)

Since, Re(z) > 0 and Im(z) < 0. So, it lies in fourth quadrant 

\(\therefore\)  arg(z) = -\(\alpha\) = -\(\pi\over 6\)

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