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How To Find Argument Of Complex Number

Written By Monday, March 14, 2022 Add Comment Edit

In this explainer, we volition learn how to place the argument of a complex number and how to calculate it.

When we plot complex numbers on an Argand diagram, nosotros can meet that complex numbers share many properties with vectors. For instance, adding and subtracting complex numbers is geometrically equivalent to the corresponding operations of vectors. We know that the characteristics of a vector are its direction and magnitude, and then a complex number must take equivalent characteristics. Nosotros recall that the magnitude of a circuitous number is called its modulus. The direction of a complex number in the Argand diagram is the statement of the complex number.

Definition: Statement of a Complex Number

The argument of a complex number is the bending, in radians, betwixt the positive real axis in an Argand diagram and the line segment between the origin and the complex number, measured counterclockwise. The statement is denoted a r g ( 𝑧 ) , or A r g ( 𝑧 ) .

The statement πœƒ of a complex number is, by convention, given in the range − πœ‹ < πœƒ ≤ πœ‹ . However, we tin likewise discuss a complex number with an argument greater than πœ‹ or less than − πœ‹ . The argument of a complex number within the range ] − πœ‹ , πœ‹ ] is chosen the principal argument. Other conventions use the range 0 ≤ πœƒ < 2 πœ‹ for the principal argument, but this is less common.

If we are given the Cartesian form, π‘Ž + 𝑏 𝑖 , of a circuitous number, we can use correct triangle trigonometry to find the argument of the complex number. For example, consider the circuitous number given in the Argand diagram above. Since this complex number lies in the first quadrant, nosotros can see that the argument of this complex number is an angle in the right triangle whose sides are the blue, greenish, and majestic line segments. In this case, the tangent of this angle is the ratio o p p o s i t e a d j a c e northward t ; hence, t a n πœƒ = 𝑏 π‘Ž .

We tin can and then compute πœƒ by applying the inverse tangent function to both sides of this equation: πœƒ = 𝑏 π‘Ž . t a northward

This method can be used whenever a complex number lies in the outset quadrant. In our offset example, nosotros will find the principal argument of a complex number in the first quadrant by using right triangle trigonometry.

Example 1: Finding the Argument of a Complex Number in Radians

Discover the statement of the complex number four + 3 𝑖 in radians. Give your answer right to two decimal places.

Respond

Recall that the argument of a complex number is the angle, in radians, between the positive real axis of an Argand diagram and the line betwixt the origin and the complex number, measured counterclockwise. We also remember that the statement of a complex number is, by convention, given in the range ] − πœ‹ , πœ‹ ] .

We begin by plotting the complex number on an Argand diagram.

Nosotros have labeled the argument of the complex number in the Argand diagram above πœƒ . We tin can run across that the argument of this circuitous number is an angle in the correct triangle whose sides are the blue, light-green, and royal line segments. Applying right triangle trigonometry, we obtain t a n o p p o s i t e a d j a c e n t πœƒ = = 3 iv .

We can and then use the inverse tangent office to both sides of this equation to find πœƒ = 3 4 = 0 . 6 four 3 5 … . a r c t a n r a d i a n s

Hence, a r chiliad r a d i a n s ( 4 + 3 𝑖 ) = 0 . six 4 to two decimal places.

In the previous example, we were able to calculate the argument of a complex number, π‘Ž + 𝑏 𝑖 , past evaluating the inverse tangent of 𝑏 π‘Ž . Even so, this is non the case for all complex numbers every bit the next example will demonstrate.

Example ii: Finding the Master Argument of a Complex Number

Given that 𝑍 = − one 2 + √ 3 2 𝑖 , find the principal statement of 𝑍 .

Answer

Recall that the argument of a complex number is the angle, in radians, betwixt the positive real centrality of an Argand diagram and the line between the origin and the complex number, measured counterclockwise. Additionally, we retrieve that the principal statement of a complex number is the statement that lies in the range ] − πœ‹ , πœ‹ ] .

We begin by plotting the circuitous number on an Argand diagram as shown below.

We have labeled the argument of the complex number in the Argand diagram above πœƒ and the supplementary angle πœ™ . Nosotros can encounter that πœ™ is an angle in the right triangle whose sides are the bluish, green, and purple line segments. Applying right triangle trigonometry, nosotros obtain t a n o p p o s i t east a d j a c east n t πœ™ = = . √

We tin and so apply the inverse tangent function to both sides of this equation to detect πœ™ = ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = √ 3 = πœ‹ three . a r c t a due north a r c t a north r a d i a n s √

We can then calculate the statement past subtracting πœ™ from πœ‹ : a r g r a d i a n south ( 𝑍 ) = πœ‹ − πœ™ = πœ‹ − πœ‹ three = 2 πœ‹ three .

We note that this argument lies in the range ] − πœ‹ , πœ‹ ] ; hence, it is the chief argument.

Nosotros conclude that the principal argument of the given circuitous number is 2 πœ‹ 3 .

In the previous example, we saw that the argument of a circuitous number, π‘Ž + 𝑏 𝑖 , is non always equal to the changed tangent of 𝑏 π‘Ž . In fact, if we had naively tried to calculate the statement of 𝑧 by evaluating 𝛼 = ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ , a r c t a n √ we would take ended upwardly with 𝛼 = − √ 3 = − πœ‹ 3 . a r c t a north r a d i a n southward

This statement represents a clockwise angle from the positive real axis of πœ‹ three radians, which would place the circuitous number in the fourth quadrant. Information technology is apparent from the Argand diagram in the previous example that this is not the argument of the circuitous number. However, we tin can arrive at the right value of a r grand ( 𝑧 ) past calculation πœ‹ to 𝛼 .

This effect demonstrates that we need to be careful when computing the argument of a complex number that does not lie in the first quadrant. Also, nosotros tin come across that there are different approaches to obtain a r g ( 𝑧 ) .

We outline two different methods of calculating the statement of a circuitous number. Whichever method we cull to use, plotting the number on an Argand diagram will be extremely useful and will assistance us avoid mutual errors in calculating the argument.

How To: Finding the Statement of a Complex Number Using the Inverse Tangent Office

To find the statement, a r 1000 ( 𝑧 ) , of a complex number 𝑧 = π‘Ž + 𝑏 𝑖 , nosotros need to consider which quadrant information technology lies in. The statement of a complex number 𝑧 = π‘Ž + 𝑏 𝑖 tin can be obtained using the inverse tangent part in each quadrant as follows:

  • If 𝑧 lies in the starting time or the quaternary quadrant, a r yard a r c t a north ( 𝑧 ) = 𝑏 π‘Ž .
  • If 𝑧 lies in the second quadrant, a r thou a r c t a northward ( 𝑧 ) = 𝑏 π‘Ž + πœ‹ .
  • If 𝑧 lies in the third quadrant, a r thousand a r c t a north ( 𝑧 ) = 𝑏 π‘Ž − πœ‹ .

If the complex number does not lie on a quadrant, and so it is either purely real or purely imaginary. If information technology is purely imaginary ( π‘Ž = 0 ) , then a r k f o r a r k f o r ( 𝑧 ) = πœ‹ 2 𝑏 > 0 , ( 𝑧 ) = − πœ‹ two 𝑏 < 0 .

If purely existent ( 𝑏 = 0 ) , then a r one thousand f o r a r g f o r ( 𝑧 ) = 0 π‘Ž > 0 , ( 𝑧 ) = πœ‹ π‘Ž < 0 .

Lastly, if π‘Ž = 𝑏 = 0 , the argument is undefined.

These points are summarized in the post-obit diagram.

The primary benefit of the method described above is that we are given a formula to follow for each situation. However, this method also requires us either to memorize each dominion or to have an bachelor reference for the rules. An alternative method for finding the argument of a complex number is to use correct triangle trigonometry to first identify the positive acute angle betwixt the existent axis and the line segment betwixt the origin and the complex number in an Argand diagram. After finding the positive astute angle, we tin can find the argument of the complex number geometrically.

How To: Finding the Argument of a Complex Number Using Positive Astute Angles

We define the angle πœƒ to be the positive acute angle between the line linking 𝑧 to the origin and the real axis every bit shown in the diagram.

We can and then summate the argument of 𝑧 in different quadrants equally follows:

  • Quadrant 1: a r g ( 𝑧 ) = πœƒ
  • Quadrant ii: a r g ( 𝑧 ) = πœ‹ − πœƒ
  • Quadrant three: a r m ( 𝑧 ) = πœƒ − πœ‹
  • Quadrant 4: a r g ( 𝑧 ) = − πœƒ

The ii unlike methods for obtaining the argument of a complex number will lead to the same answer. The second method, which uses the positive acute angle, is more intuitive and requires less memorization. Using this method, we first compute the positive acute angle and so use it to discover the argument of the circuitous number, which is the counterclockwise angle from the positive real axis, lying in the range ] − πœ‹ , πœ‹ ] .

In the next example, we will utilize this method to find the argument of a complex number lying in the third quadrant.

Example 3: The Relationship betwixt the Complex Conjugate and the Argument

Consider the complex number 𝑧 = − four − 5 𝑖 .

  1. Calculate a r g ( 𝑧 ) , giving your reply correct to two decimal places in an interval from − πœ‹ to πœ‹ .
  2. Summate a r g 𝑧 , giving your answer right to two decimal places in an interval from − πœ‹ to πœ‹ .

Answer

Recall that the argument of a complex number is the bending, in radians, betwixt the positive real axis of an Argand diagram and the line between the origin and the circuitous number, measured counterclockwise. We besides remember that the statement of a complex number is, past convention, given in the range ] − πœ‹ , πœ‹ ] .

Part ane

We brainstorm by plotting the complex number on an Argand diagram equally shown beneath.

Nosotros have labeled the acute bending πœ™ , which is related to the argument of the complex number 𝑧 . If we can find angle πœ™ , the statement of this number tin exist obtained by calculation πœ‹ to this angle. However, we tin can see that this statement will non prevarication in the range ] − πœ‹ , πœ‹ ] . We must and so decrease the full revolution ii πœ‹ from this resulting angle, which leads to the relation a r g ( 𝑧 ) = ( πœ™ + πœ‹ ) − ii πœ‹ = πœ™ − πœ‹ .

We tin can see that πœ™ is an angle in the right triangle whose sides are the blue, light-green, and royal line segments. Applying right triangle trigonometry, we obtain t a due north o p p o s i t due east a d j a c e due north t πœ™ = = 5 iv .

We can then utilize the changed tangent function to both sides of this equation to find πœ™ = five 4 = 0 . 8 9 6 0 … . a r c t a n r a d i a n south

Hence, to calculate a r yard ( 𝑧 ) , we subtract πœ‹ from πœ™ , which gives a r g r a d i a northward s r o u n d e d t o d e c i k a l p l a c e s ( 𝑧 ) = πœ™ − πœ‹ = − 2 . two 4 5 five … = − 2 . 2 5 , 2 .

Office 2

Recall that the cohabit 𝑧 is obtained by switching the sign of the imaginary part of the complex number 𝑧 . Hence, 𝑧 = − 4 + five 𝑖 . We now plot 𝑧 on an Argand diagram.

Similar to the previous part, we volition find the statement of 𝑧 past beginning calculating πœ™ : πœ™ = v 4 = 0 . viii 9 6 0 … . a r c t a n r a d i a n s

Since πœ™ and a r g 𝑧 are supplementary, we tin can obtain a r thou 𝑧 by subtracting πœ™ from πœ‹ : a r thousand r a d i a n s r o u n d eastward d t o d east c i m a l p l a c e s 𝑧 = πœ‹ − πœ™ = ii . 2 4 v 5 … = 2 . ii v , two .

In the previous case, we computed the arguments of a complex number and its conjugate. Nosotros tin can notation that the argument of the circuitous conjugate in this instance is the negative of the argument of the original complex number. This demonstrates a general rule of the argument.

Property: Argument of the Conjugate of a Complex Number

Given any nonzero complex number 𝑧 and its conjugate 𝑧 (also denoted 𝑧 ∗ ), a r chiliad a r chiliad ( 𝑧 ) = − 𝑧 .

In the next case, nosotros volition demonstrate how the multiplication and division of complex numbers is associated with the arguments of the complex numbers.

Example 4: Arguments of Products and Quotients

Consider the complex numbers 𝑧 = 1 + √ 3 𝑖 and 𝑀 = 2 − 2 𝑖 .

  1. Detect a r g ( 𝑧 ) and a r g ( 𝑀 ) .
  2. Summate a r g ( 𝑧 𝑀 ) . How does this compare to a r g ( 𝑧 ) and a r g ( 𝑀 ) ?
  3. Calculate a r g 𝑧 𝑀 . How does this compare to a r g ( 𝑧 ) and a r g ( 𝑀 ) ?

Answer

Recall that the statement of a complex number is the angle, in radians, between the positive real axis of an Argand diagram and the line betwixt the origin and the complex number, measured counterclockwise. We also recollect that the argument of a circuitous number is, by convention, given in the range ] − πœ‹ , πœ‹ ] .

Part 1

Allow united states of america start by plotting 𝑧 and 𝑀 on an Argand diagram.

Recall that the argument of a complex number 𝑧 = π‘Ž + 𝑏 𝑖 lying in the showtime or the fourth quadrant is given by a r 1000 a r c t a n ( 𝑧 ) = 𝑏 π‘Ž .

Since 𝑧 and 𝑀 lie in the first and fourth quadrants, respectively, nosotros tin use the changed tangent to detect their arguments as follows: a r g a r c t a n r a d i a n due south ( 𝑧 ) = √ 3 1 = πœ‹ 3 and a r g a r c t a due north r a d i a north south ( 𝑀 ) = − two 2 = − πœ‹ 4 .

Function ii

We begin past calculating 𝑧 𝑀 as follows: 𝑧 𝑀 = one + √ 3 𝑖 ( 2 − 2 𝑖 ) .

Multiplying through the parenthesis, we obtain 𝑧 𝑀 = ii − 2 𝑖 + 2 √ 3 𝑖 − 2 𝑖 √ 3 .

Using 𝑖 = − ane and gathering real and imaginary terms, we obtain 𝑧 𝑀 = 2 + 2 √ 3 + two √ 3 − two 𝑖 .

Since both the existent and the imaginary parts are positive, 𝑧 𝑀 lies in the first quadrant of the Argand diagram and nosotros can summate the argument by evaluating the inverse tangent equally follows: a r chiliad a r c t a n ( 𝑧 𝑀 ) = 2 √ 3 − 2 2 + 2 √ 3 .

Canceling the cistron 2 from the top and the bottom, we take a r chiliad a r c t a n ( 𝑧 𝑀 ) = √ 3 − 1 1 + √ 3 .

We can simplify the fraction by multiplying both the numerator and the denominator by the cohabit of the denominator: a r yard a r c t a n ( 𝑧 𝑀 ) = √ 3 − ane 1 − √ three one + √ three one − √ 3 .

Multiplying through the parenthesis, we obtain a r grand a r c t a due north a r c t a n a r c t a n r a d i a n south ( 𝑧 𝑀 ) = − i + 2 √ 3 − 3 1 − three = − 4 + ii √ 3 − 2 = ii − √ three = πœ‹ ane 2 .

Comparing this with a r g ( 𝑧 ) and a r chiliad ( 𝑀 ) , nosotros observe that a r yard a r g a r g ( 𝑧 𝑀 ) = ( 𝑧 ) + ( 𝑀 ) .

Part 3

We start by computing 𝑧 𝑀 as follows: 𝑧 𝑀 = 1 + √ three 𝑖 2 − two 𝑖 .

To write this complex number in the Cartesian form, π‘Ž + 𝑏 𝑖 , we demand to multiply both the numerator and the denominator by the conjugate of the denominator, which is 2 + 2 𝑖 : 𝑧 𝑀 = one + √ 3 𝑖 ( ii + 2 𝑖 ) ( 2 − 2 𝑖 ) ( 2 + 2 𝑖 ) .

Multiplying through the parenthesis, we have 𝑧 𝑀 = two + 2 𝑖 + 2 𝑖 √ three + 2 𝑖 √ 3 4 + 4 .

Using 𝑖 = − i and gathering real and imaginary terms, 𝑧 𝑀 = 1 4 1 − √ 3 + 1 4 1 + √ iii 𝑖 .

Since R east 𝑧 𝑀 < 0 and I m 𝑧 𝑀 > 0 , the complex number 𝑧 𝑀 lies in the 2d quadrant. Recall that if a complex number 𝑧 = π‘Ž + 𝑏 𝑖 lies in the second quadrant, a r g a r c t a n ( 𝑧 ) = 𝑏 π‘Ž + πœ‹ .

Hence, nosotros have a r g a r c t a north 𝑧 𝑀 = 1 + √ 3 1 − √ 3 + πœ‹ .

Canceling the common factor ane iv , we have a r g a r c t a n 𝑧 𝑀 = ane + √ 3 one − √ 3 + πœ‹ .

Evaluating the inverse tangent, we obtain a r k 𝑧 𝑀 = − five πœ‹ 1 2 + πœ‹ = 7 πœ‹ i ii .

Finally, comparing this with a r thousand ( 𝑧 ) and a r k ( 𝑀 ) , we find that a r g a r yard a r g 𝑧 𝑀 = ( 𝑧 ) − ( 𝑀 ) .

In the previous example, we observed the relationship betwixt the multiplication/division of circuitous numbers and their arguments. This relationship shown in the instance holds for full general complex numbers.

Property: Arguments and Multiplication/Division of Complex Numbers

Given whatsoever nonzero complex numbers 𝑧 and 𝑧 , a r thousand a r thousand a r chiliad a r g a r g a r g ( 𝑧 𝑧 ) = ( 𝑧 ) + ( 𝑧 ) , 𝑧 𝑧 = ( 𝑧 ) − ( 𝑧 ) .

The next instance volition demonstrate how we can solve bug by applying the backdrop of the statement.

Example 5: Using Multiplication of Circuitous Numbers to Determine an Statement

A complex number is multiplied past another complex number 𝑧 and so past the complex conjugate 𝑧 . How is the statement of the resulting complex number related to the statement of the original complex number?

Reply

Recall that the argument of the production of a pair of complex numbers is equal to the sum of the arguments of the two circuitous numbers.

We kickoff with a complex number 𝑀 ; and then, it is multiplied past 𝑧 and 𝑧 . Hence, the result is 𝑀 𝑧 𝑧 . We are asked how the argument of the resulting complex number is related to the argument of the original circuitous number. Therefore, we should consider a r g 𝑀 𝑧 𝑧 . Using multiplicative properties of the argument, we can rewrite this as follows: a r g a r g a r g a r chiliad 𝑀 𝑧 𝑧 = ( 𝑀 ) + ( 𝑧 ) + 𝑧 .

Nosotros also know that the argument of a complex number equals the negative of the argument of its conjugate. Hence, we tin supervene upon a r g 𝑧 above with − ( 𝑧 ) a r g to write a r thousand a r g a r g a r one thousand a r 1000 𝑀 𝑧 𝑧 = ( 𝑀 ) + ( 𝑧 ) − ( 𝑧 ) = ( 𝑀 ) .

Therefore, the statement of the complex number after it is multiplied by another complex number 𝑧 and and then by the complex conjugate 𝑧 is unchanged.

In our final example, we will consider the relationship between the argument and powers.

Example vi: Finding the Argument of Powers of Complex Numbers in Algebraic Form

Consider the complex number 𝑧 = 7 + 7 𝑖 .

  1. Find the statement of 𝑧 .
  2. Hence, discover the statement of 𝑧 .

Answer

Recall that the argument of a complex number is the angle, in radians, betwixt the positive real axis of an Argand diagram and the line between the origin and the complex number, measured counterclockwise. Nosotros also retrieve that the statement of a complex number is, by convention, given in the range ] − πœ‹ , πœ‹ ] .

Role one

Recollect that the statement of a circuitous number 𝑧 = π‘Ž + 𝑏 𝑖 lying in the first or the fourth quadrant is given past a r one thousand a r c t a n ( 𝑧 ) = 𝑏 π‘Ž .

Since the circuitous number lies in the first quadrant, nosotros can summate its argument past evaluating the inverse tangent of its imaginary part over its existent role as follows: a r 1000 a r c t a n a r c t a n r a d i a north due south ( 𝑧 ) = 7 seven = ( 1 ) = πœ‹ 4 .

Part 2

We recall that for any two nonzero complex numbers 𝑧 and 𝑧 , a r chiliad a r g a r m ( 𝑧 𝑧 ) = ( 𝑧 ) + ( 𝑧 ) .

If the circuitous numbers are both equal to 𝑧 , this means a r g a r chiliad 𝑧 = 2 ( 𝑧 ) .

Using similar logic, we tin find that a r thousand a r thousand a r g a r g 𝑧 = 3 ( 𝑧 ) , 𝑧 = 4 ( 𝑧 ) .

Therefore, a r g a r yard r a d i a due north s 𝑧 = 4 ( 𝑧 ) = iv × πœ‹ four = πœ‹ .

In the previous instance, nosotros saw the relationship between the power of a complex number and its statement. Using similar logic equally in that example, we can see that this relationship holds for general complex numbers for whatsoever positive integer power.

Property: Statement of the Ability of a Complex Number

Given any nonzero complex number 𝑧 and a positive integer ability 𝑛 , the argument of 𝑧 is given by a r grand a r yard ( 𝑧 ) = 𝑛 ( 𝑧 ) .

In this explainer, we accept considered how the argument of a circuitous number relates to the conjugates, multiplication, and partition of complex numbers. However, we take intentionally left out the add-on and subtraction of circuitous numbers since at that place is no simple relationship between these operations and the arguments of complex numbers. We will finish this explainer past illustrating in two different ways why we exercise not expect to observe a elementary relationship betwixt add-on/subtraction and the arguments of complex numbers.

Firstly, nosotros recollect that the addition and subtraction of complex numbers are geometrically equivalent to the corresponding vector operations and hence follow the triangle or parallelogram rules. In doing so, nosotros can run into that knowing only the arguments (angles) of the complex numbers volition non be sufficient to find the argument of the resulting circuitous number. This is one manner that we tin can see why there is no simple relationship between these operations and the arguments of complex numbers.

As an alternative way to meet why no such simple human relationship exists, let the states consider the iii complex numbers 𝑧 = ane + 𝑖 , 𝑧 = 2 + √ 3 ( 1 + 𝑖 ) , and 𝑧 = 1 − 𝑖 plotted on the Argand diagram below.

We tin encounter that a r one thousand a r thou ( 𝑧 ) = ( 𝑧 ) = πœ‹ 4 and that a r g ( 𝑧 ) = − πœ‹ 4 . Furthermore, 𝑧 + 𝑧 = 2 , which has an argument of 0, whereas 𝑧 + 𝑧 = 3 + √ three + 1 + √ iii 𝑖 , whose argument is clearly not zero. We can, in fact, calculate the exact value of the argument equally follows: a r g a r c t a n ( 𝑧 + 𝑧 ) = 1 + √ 3 3 + √ three .

By multiplying both the numerator and the denominator by the conjugate of the denominator, nosotros can simplify the fraction: a r thou a r c t a n ( 𝑧 + 𝑧 ) = 1 + √ iii 3 − √ 3 3 + √ 3 3 − √ iii .

Multiplying through the parenthesis, we obtain a r thou a r c t a n ( 𝑧 + 𝑧 ) = iii − √ three + 3 √ 3 − 3 iii − 3 .

Finally, we can simplify and evaluate the inverse tangent to obtain a r g a r c t a n a r c t a n r a d i a n southward ( 𝑧 + 𝑧 ) = 2 √ 3 half-dozen = √ iii three = πœ‹ six .

To summarize what we take calculated hither, recall that the complex numbers 𝑧 and 𝑧 have the same argument, πœ‹ four . If a simple relationship between the arguments of circuitous numbers and the sum existed, the arguments of 𝑧 + 𝑧 and 𝑧 + 𝑧 would be the same. However, nosotros have obtained a r yard a r k ( 𝑧 + 𝑧 ) = 0 , ( 𝑧 + 𝑧 ) = πœ‹ 6 .

This demonstrates that knowing the arguments of two complex numbers is not sufficient to be able to calculate the argument of their sum.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • The argument of a circuitous number 𝑧 is defined as the angle, in radians, between the positive real axis in an Argand diagram and the line segment from the origin to the complex number, measured counterclockwise.
  • The statement of a complex number 𝑧 = π‘Ž + 𝑏 𝑖 can be obtained using the inverse tangent function in each quadrant equally follows:
    • If 𝑧 lies in the get-go or the fourth quadrant, a r g a r c t a n ( 𝑧 ) = 𝑏 π‘Ž .
    • If 𝑧 lies in the 2d quadrant, a r one thousand a r c t a northward ( 𝑧 ) = 𝑏 π‘Ž + πœ‹ .
    • If 𝑧 lies in the third quadrant, a r yard a r c t a n ( 𝑧 ) = 𝑏 π‘Ž − πœ‹ .
  • The statement has the following properties:
    • a r g a r g ( 𝑧 ) = − 𝑧 ,
    • a r grand a r g a r g ( 𝑧 𝑧 ) = ( 𝑧 ) + ( 𝑧 ) ,
    • a r 1000 a r thou a r 1000 𝑧 𝑧 = ( 𝑧 ) − ( 𝑧 ) ,
    • a r thousand a r g ( 𝑧 ) = 𝑛 ( 𝑧 ) .
  • There is no simple relationship between the add-on of complex numbers and their arguments.

Source: https://www.nagwa.com/en/explainers/184185851978/

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