The complex plane represents a geometric interpretation of complex numbers. This plane, which is equivalent to the Cartesian plane, represents the real and imaginary components of a complex number, as well as the \(X\) and \(Y\) axes. The magnitude and argument of a complex number are required for the representation of any complex number. The complex plane is very important in maths.

It’s also called the Argand plane because it’s made up of two mutually perpendicular axes. The horizontal line that represents real numbers is known as the real axis. The vertical line, on the other hand, represents imaginary numbers and is referred to as an imaginary axis. Complex numbers can be referred to as the extension of the one-dimensional number line. In this article, let’s understand everything about the argument of a complex number in detail.

Define Argument of a Complex Number

The angle inclined from the real axis in the direction of the complex number displayed on the complex plane is defined as the argument of a complex number.

1. Denoted By: Argument of a complex number is denoted by \(“\,\,\theta \,\,”\,or\,\,\,\varphi \,\)
2. Measurement: Argument of a complex number is measured in standard units called Radians

Origin of Complex Number

Now that we understood the definition of the argument of a complex number, let’s understand its origin in brief.

Complex numbers are the numbers that can be written in the form of \(x + iy,\) where \(x,y\) are real numbers and \(i = \sqrt { – 1} \) Here, \(i\) is an imaginary number whose square is \( – 1\)

If \(z = a + ib\) is the complex number, then \(a\) and \(b\) are called real and imaginary parts, respectively, of the complex number and written as \({\mathop{\rm Re}\nolimits} (z) = a,{\mathop{\rm Im}\nolimits} \,(z)\, = \,b\)

A complex number \(z\, = \,a\, + ib\) can be represented by a unique point \(P(a,b)\) in the cartesian plane referred to as a pair of rectangular axes. The complex number \(0 + 0i\) represents the origin \(O(0,0).\) A purely real number \(a\) , i.e.,\((a\, + \,0i)\,\) is represented by the point \((a,0)\) on \(x\)-axis. Therefore, the \(x\)-axis is called a real axis. A purely imaginary number \(ib,i.e.,(0 + ib)\) is represented by the point on \(y\)-axis Therefore, is called an imaginary axis.

Similarly, the representation of complex numbers as points in the plane is known as the Argand diagram. The Argand plane, often known as the complex plane, is a plane that represents complex numbers as points.

Let \(P\) be a point representing a non-zero complex number \(z = a + ib\) in the Argand plane. If \(OP\) makes an angle \(\theta \) with the positive direction of \(x\)-axis , then \(z = r(\cos \,\theta  + i\sin \theta )\) is called the polar form of the complex number, where
\(r = \left| z \right| = \sqrt {{a^2} + {b^2}} \) and \(\tan \,\theta  = \frac{b}{a}.\) Here \(\theta\) is called argument or amplitude of \(z\) is represented by \(\arg (z) = \theta .\) It is generally measured in radians. So, The general argument of complex number \(z\) is represented by \(\arg (z) = \theta  + 2n\pi \) where \(n\) is an integer.

In the above diagram, we can see a complex number \(z = x + iy = P(x,y)\) is represented as a point . The angle made by a line joining \(P\) to the origin from the positive real axis is called the argument of point \(P\) and the length of the line from \(P\) to the origin is called the modulus of complex number \(z.\)

What is the Principal Argument of a Complex Number?

The argument of a complex number \(z = a + ib\) is the angle \(\theta \) of its polar representation. This angle is multi-valued. If \(\theta \) is the argument of a complex number \(z\),then \(\theta  + 2n\pi \) will also be argument of that complex number, where \(n\) is an integer.
While, principal argument of a complex number is the unique value of \(\theta \) such that \( – \pi  < \theta  \le \pi .\)

So, the principal argument of a complex number is always a unique data point, while argument of a complex number has multiple data points due to its integral multiple of \(2\pi \)

For example:
\(z = i = P(0,1)\) which lie on the positive imaginary axis; hence argument of \(z\,is\,\frac{\pi }{2} + 2n\pi \)
But its Principal argument will be only \(\frac{\pi }{2}\).

Argument of a Complex Number Formula

Let \(P\) be a point representing a non-zero complex number \(z = a + ib\) in the Argand plane. If \(OP\) makes an angle \(\theta \) with the positive direction of \(x\)-axis then \(z = r(\cos \,\,\theta  + \,i\sin \,\theta )\) is called the polar form of the complex number, where \(r = \left| z \right| = \sqrt {{a^2} + {b^2}}\) and \(\tan \,\theta \, = \frac{b}{a}\) is called argument or amplitude of \(z\) and we write it as arg \((z)\, = \,\theta. \) The general argument of a complex number is represented by \(\theta  + 2n\pi \) where \(n\) is an integer.

In short, \(\arg \,(z) = \,\theta  = \arg (a + ib) = {\tan ^{ – 1}}\frac{b}{a}\)

Example 1: Find the argument of a complex number \(1 + i\)
Solution: \(\arg \,(z) = \,\theta  = \arg (1 + i1) = {\tan ^{ – 1}}\frac{1}{1} = \frac{\pi }{4}\)(since, point will be in \({1^{st}}\)quadrant)
Hence, general argument of this complex number is \(\frac{\pi }{4} + 2n\pi.\)

Example 2: Find the argument of a complex number \( – 1 + i\)
Solution: \(\arg \,(z) = \,\theta  = \arg ( – 1 + i1) = {\tan ^{ – 1}}\frac{{ – 1}}{1} = \frac{{3\pi }}{4}\) (since, point will be in \({2^{nd}}\)quadrant)
Hence, general argument of this complex number is \(\frac{{3\pi }}{4} + 2n\pi.\)

Example 3: Find the argument of a complex number \( – 1 – i\)
Solution: \(\arg (z) = \theta  = \arg ( – 1 + i( – 1)) = {\tan ^{ – 1}}\frac{{ – 1}}{{ – 1}} = \frac{{5\pi }}{4}\) (since, point will be in \({3^{rd}}\) quadrant)
Hence, general argument of this complex number is \(\frac{{5\pi }}{4} + 2n\pi.\)
But if someone ask you about the principal argument then it will be \( – \frac{{3\pi }}{4}\) (putting \(n{\rm{ }} = \;\, – {\rm{ }}1\) in the general argument so that angle will be in interval \( – \pi  < \theta  \le \pi \) )

Example 4: Find the argument of a complex number \( 1 – i\)
Solution: \(\arg (z) = \theta  = \arg ( 1 + i( – 1)) = {\tan ^{ – 1}}\frac{-1}{{ 1}} = \frac{{7\pi }}{4}\) (since, point will be in \({4^{th}}\) quadrant)
Hence, general argument of this complex number is \(\frac{{7\pi }}{4} + 2n\pi.\)
But if someone ask you about the principal argument then it will be \( – \frac{\pi }{4}\) (putting \(n{\rm{ }} = \;\, – {\rm{ }}1\) in the general argument so that angle will be in interval \( – \pi  < \theta  \le \pi \)

Properties of Argument of a Complex Number

The properties of argument of a complex number are as follows:
If \(z,\,{z_1},\,{z_2}\) be non-zero complex number and \(m\) be any integer, then

1. \(\arg ({z_1}\,.\,{z_2})\, = \,\arg \,({z_1}) + \arg ({z_2})\)
2. \(\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \arg ({z_1}) – \arg ({z_2})\)
3. \(\arg ({z^m}) = {\rm{m}}\,{\rm{arg}}({\rm{z}})\)

Solved Examples of Argument of a Complex Number

Q.1: Let \({{\rm{z}}_1}\,and\,{z_2}\)be two complex numbers such that \({\overline z _1}\, + i{\overline z _2}\, = \,0\) and \(\arg ({z_1}{z_2}) = \pi \). Then find \(\arg ({z_1})\)
Ans: Given that \(\overline {{z_1}} + i\overline {{z_2}} = 0 \Rightarrow {z_1} = i{z_2},\) i.e., \({z_2} = – i{z_1}\)
Thus \(\arg ({z_1}{z_2}) = \arg \,{z_1} + \arg \,( – i{z_1}) = \pi \)
\(\Rightarrow {\mkern 1mu} {\mkern 1mu} \arg \left( {iz_1^2} \right) = \pi\)
\(\Rightarrow {\mkern 1mu} \arg \left( { – i} \right) + \arg \left( {z_1^2} \right) = \pi\)
\(\Rightarrow {\mkern 1mu} \arg \left( { – i} \right){\mkern 1mu} + 2\arg \left( {{z_1}} \right) = \pi\)
\(\Rightarrow \frac{{ – \pi }}{2} + 2\arg \left( {{z_1}} \right) = \pi\)
\(\Rightarrow \arg {\mkern 1mu} \left( {{z_1}} \right) = \frac{{3\pi }}{4}\)

Q.2: Let \({z_1}\) and \(\,{z_2}\) be two complex numbers such that \(\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|\) then show that \(\arg ({z_1}) – \arg ({z_2}) = 0.\)
Ans: Let \({z_1} = {r_1}\left( {\cos {\mkern 1mu} {\theta _1} + i\sin {\mkern 1mu} {\theta _1}} \right)\) and \({z_2} = {r_2}\left( {\cos {\mkern 1mu} {\mkern 1mu} {\theta _2} + i\sin {\mkern 1mu} {\theta _2}} \right)\)
Where \({r_1} = \left| {{z_1}} \right|,\arg ({z_1}) = {\theta _1},{r_2} = \left| {{z_2}} \right|,\arg ({z_2}) = {\theta _2}.\)
We have
\(\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|\)
\( = \left| {{r_1}(\cos \,{\theta _1} + \cos \,\,{\theta _2}) + {r_2}(\cos \,{\theta _2} + \sin \,{\theta _2})} \right| = {r_1} + {r_2}\)
\(= r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {{\theta _1} – {\theta _2}} \right) = {\left( {{r_1} + {r_2}} \right)^2} \Rightarrow \cos \left( {{\theta _1} – {\theta _2}} \right) = 1\)
\( \Rightarrow {\theta _1}\, – \,{\theta _2}\,i.e.\arg \,{z_1} = \arg \,{z_2}\)

Q.3: If \({z_1}\) and \({z_2}\) both satisfy \(z{\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} \bar z{\mkern 1mu} = 2\left| {z – 1} \right|,\arg \left( {{z_1} – {z_2}} \right) = \frac{\pi }{4},\) then find \({\rm{Im}}\left( {{z_1} + {z_2}} \right)\)
Ans: Let \(z = x + iy,{z_1} = {x_1} + i{y_1}\) and \({z_2} = {x_2} + i{y_2}\)
Then \(z\, + \,\,\overline z \, = 2\left| {z – 1} \right|\)
\(\Rightarrow \left( {x + iy} \right) + \left( {x – iy} \right) = 2\left| {x – 1 + iy} \right|\)
\(\Rightarrow 2x = 1 + {y^2}\,\,\,\,…(1)\)
Since \({z_1}\) and \({z_2}\) both satisfy \((1),\) we have
\(2{x_1} = 1 + y_1^2\) and \(2{x_2} = 1 + y_2^2\)
\(\Rightarrow 2\left( {{x_1} – {\mkern 1mu} {x_2}} \right) = \left( {{y_1} + {y_2}} \right)\left( {{y_1} – {y_2}} \right)\)
\(\Rightarrow 2 = \left( {{y_1} + {y_2}} \right)\left( {\frac{{{y_1} – {y_2}}}{{{x_2} – {x_2}}}} \right)\,\,…(2)\)
Again \({z_1} – {z_2} = \left( {{x_1} – {x_2}} \right) + i\left( {{y_1} – {y_2}} \right)\)
Therefore \(\tan \,\theta  = \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}},\) where \(\theta = \arg \left( {{z_1} – {z_2}} \right)\)
\(\Rightarrow \tan \frac{\pi }{4} = \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}\) (since \(\theta  = \frac{\pi }{4}\))
i.e.,\(1 = \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}\)
From \((2)\) we get \(2 = {y_1} + {y_2}\) i.e., \({\mathop{\rm Im}\nolimits} ({z_1} + {z_2}) = 2\)

Q.4: If \(\arg (z – 1) = \arg (z + 3i),\) then find \((x – 1):y\) where \(z = x + iy\)
Ans: we have \(\arg (z – 1) = \arg (z + 3i),\) where \(z = x + iy\)
\( \Rightarrow \,\arg \,(x + iy – 1) = \arg (x + iy + 3i)\)
\( \Rightarrow \arg (x – 1 + iy) = \arg \left[ {x + i(y + 3)} \right]\)
\( \Rightarrow {\tan ^{ – 1}}\frac{y}{{x – 1}} = {\tan ^{ – 1}}\frac{{y + 3}}{x} \Rightarrow \frac{y}{{x – 1}} = \frac{{y + 3}}{x}\)
\( \Rightarrow xy = (x – 1)(y + 3)\)
\( \Rightarrow xy = xy – y + 3x – 3\,\, \Rightarrow 3x – 3 = y\)
\( \Rightarrow \frac{{x – 1}}{y} = \frac{1}{3}\)

Q.5: Show that the complex number \(z\) satisfying the condition \(\arg \left( {\frac{{z – 1}}{{z + 1}}} \right) = \frac{\pi }{4}\) lies on a circle.
Ans: Given \(\arg \left( {\frac{{z – 1}}{{z + 2}}} \right) = \frac{\pi }{4}\)
\(\arg \left( {\frac{{z – 1}}{{z + 2}}} \right) = \frac{\pi }{4}\)
\( \Rightarrow \arg (z – 1) – \arg (z + 1) = \frac{\pi }{4}\)
\( \Rightarrow \arg (x – 1 + iy) – \arg (x + 1 + iy) = \frac{\pi }{4}\)
\( \Rightarrow {\tan ^{ – 1}}\frac{y}{{x – 1}} – {\tan ^{ – 1}}\frac{y}{{x + 1}} = \frac{\pi }{4} \Rightarrow {\tan ^{ – 1}}\left[ {\frac{{\frac{y}{{x – 1}} – \frac{y}{{x + 1}}}}{{1 + \left( {\frac{y}{{x – 1}}} \right)\left( {\frac{y}{{x + 1}}} \right)}}} \right] = \frac{\pi }{4}\)
\( \Rightarrow \frac{{y(x + 1 – x + 1)}}{{{x^2} – 1 + {y^2}}} = \tan \frac{\pi }{4} \Rightarrow \frac{{2y}}{{{x^2} + {y^2} – 1}} = 1\)
\( \Rightarrow {x^2} + {y^2} – 1 = 2y\)
\( \Rightarrow {x^2} + {y^2} – 2y – 1 = 0\) which represents a circle.

Summary

In this article, we learnt about the Argument of a Complex Number, its formula and examples and got to know that it plays an essential role in the mathematical world. Complex numbers are impossible to ignore in the scientific world, despite the fact that humans cannot physically visualise them. Complex numbers exemplify the importance of mathematics in science, as it serves as both a powerful language for describing complex phenomena and a full toolkit for solving challenging issues.

FAQs on Argument of a Complex Number

Q.1: What are examples of arguments for a complex number?
Ans: Some examples are:
If \(z = 1 + i \Rightarrow \arg (z) = \frac{\pi }{4};\)
If \(z =  – 1 + i \Rightarrow \arg (z) = \frac{{3\pi }}{4};\)
If \(z =  – 1 – i \Rightarrow \arg (z) = \frac{{5\pi }}{4};\)
If \(z = 1 – i \Rightarrow \arg (z) = \frac{{7\pi }}{4}\)
For the general argument, we will add \(2n\pi \) in all the above arguments.

Q.2: What is the argument of complex number \(1 – i\)?
Ans: \(\arg (z) = \theta  = \arg (1 + i( – 1)) = {\tan ^{ – 1}}\frac{1}{{ – 1}} = \frac{{7\pi }}{4} + 2n\pi \) (since, point will be in \({4^{th}}\)quadrant). Hence, general argument of this complex number is \(\frac{{7\pi }}{4} + 2n\pi \) and the principal argument will be \( – \frac{\pi }{4}\) (putting \(n\, = \, – 1\) in the general argument so that angle will be in interval\( – \pi  < \theta  \le \pi \))

Q.3: How is \(\arg (z)\) calculated?
Ans: If \(z = x + iy,\) then \(\arg (z) = \theta  = \arg (x + iy) = {\tan ^{ – 1}}\frac{y}{x}.\) If \(\theta \) is the argument of a complex number \(z\) then \(\theta  + 2n\pi \) will also be argument of that complex number, where \(n\) is an integer.

Q.4: What is the modulus and argument of a complex number?
Ans: The argument of a complex number is the angle that the line joining the complex number to the origin makes with the positive direction of the real axis. So, The argument of a complex number \(z\) is represented by \(\arg (z) = \theta \) and the length of line of complex number \(z\) from the origin is called the modulus of the complex number. It is denoted by \(\left| z \right|.\)

Q.5: What are the uses of an argument of a complex number?
Ans: Complex number’s argument plays an important role in signal processing, \(AC\) circuit analysis, Quantum mechanics etc. It helps in solving difficult problems with ease. When it comes to modelling systems with sinusoidal inputs, complex numbers offer certain important mathematical qualities that can make your job easier. A good example is electrical circuits. To summarise, you use complex numbers to avoid having to learn calculus. Even the most difficult geometrical problems are easily tackled with the concept of complex numbers.

Learn Properties of Argument of Complex Numbers

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