# Question Video: Using the Modulus and Argument to Calculate Powers of Complex Numbers in Algebraic Form Mathematics

Consider the complex number π§ = 1 + β(3) π. Find the modulus of π§. Find the argument of π§. Hence, use the properties of multiplication of complex numbers in polar form to find the modulus and argument of π§Β³. Hence, find the value of π§Β³.

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### Video Transcript

Consider the complex number π§ is equal to one plus root three π. Part one, find the modulus of π§. Part two, find the argument of π§. Part three, hence, use the properties of multiplication of complex numbers in polar form to find the modulus and argument of π§ cubed. And part four, hence, find the value of π§ cubed.

In this question, weβre given a complex number π§. And weβre given four parts to our question to help us find the value of π§ cubed. And the first part of this question wants us to find the modulus of π§. To do this, we start by noting π§ is given in algebraic form. And we recall thatβs the form π plus ππ, where π and π are real numbers. π is called the real part of our complex number and π is called the imaginary part of our complex number. And we need to find the modulus of a complex number given in algebraic form.

To do this, we recall that the modulus of a complex number is its distance from the origin in an Argand diagram. And since the complex number π plus ππ given in algebraic form has coordinates π, π on an Argand diagram, the modulus of π plus ππ will be the square root of π squared plus π squared. Thatβs the square root of the sum of the squares of its real and imaginary parts. Therefore, we can use this to find the modulus of π§: our value of π is one and our value of π is root three. So the modulus of π§ is the square root of one squared plus root three squared, which gives us the square root of four, which we can calculate is equal to two. Therefore, weβve shown the modulus of π§ is equal to two.

The second part of this question wants us to determine the argument of π§. And we recall the argument of a complex number π§ is the angle the line between the origin and π§ on an Argand diagram makes with the positive real axis measured counterclockwise. And there are many different ways of finding the argument of a complex number. First, weβre going to want to determine which quadrant our complex number lies in on an Argand diagram.

Since the real part of π§ is one and the imaginary part of π§ is root three and both of these values are positive, we can see that π§ will lie in the first quadrant of our Argand diagram. We can then connect π§ to the origin, and then the argument of π§ is the angle that this line segment makes with the positive real axis measured counterclockwise. And since π§ lies in the first quadrant, the argument of π§ will be the inverse tangent of its imaginary part divided by its real part. This is because we can construct a right triangle where the argument of π§ is an angle in this right triangle. The opposite side has length imaginary part of π§. And the adjacent side has length real part of π§. Therefore, the argument of π§ is equal to the inverse tangent of root three divided by one, which we can calculate is π by three. Therefore, the argument of π§ is π by three.

The third part of this question wants us to use the properties of multiplication of complex numbers in polar form to find the modulus and argument of π§ cubed. To do this, letβs start by recalling what we mean by complex numbers given in polar form. We recall that the polar form of a complex number π§ one is the form π times cos π plus π sin of π, where π is the modulus of π§ sub one and π is the argument of π§ sub one. And in fact, in the first two parts of this question, we found the modulus of π§ and the argument of π§. This means that we can write our complex number π§ in polar form. π is equal to two, and π is π by three. π§ is two times the cos of π by three plus π sin of π by three.

We want to use the polar form of complex numbers and their property to help us find the modulus and the argument of π§ cubed. To do this, letβs recall how we multiply complex numbers given in polar form. Letβs say we have two complex numbers given in polar form: π§ sub one with modulus π sub one and argument π sub one and π§ sub two with modulus π sub two and argument π sub two. Then we can multiply the two numbers given in polar form by using the following. π§ sub one times π§ sub two is equal to π sub one times π sub two multiplied by the cos of π one plus π two plus π sin π one plus π two.

In other words, the modulus of the product of two complex numbers is the product of their moduli. And the argument of the product of two complex numbers is the sum of their arguments. We want to use this to find the modulus and argument of π§ cubed. And we can do this by noting that π§ cubed is π§ multiplied by π§ multiplied by π§. First, the modulus of π§ cubed is going to be the modulus of π§ all cubed. And we found in the first part of the question the modulus of π§ is two. So the modulus of π§ cubed is two cubed, which we can calculate is equal to eight.

Next, letβs determine the argument of π§ cubed. Remember, when we multiply complex numbers together, we add their arguments together. And since π§ cubed is π§ multiplied by π§ multiplied by π§, the argument of π§ cubed will be the argument of π§ plus the argument of π§ plus the argument of π§ or three times the argument of π§. And in the second part of this question, we showed the argument of π§ is π by three. Therefore, the argument of π§ cubed is three times π by three, which is equal to π. Therefore, weβve shown the modulus of π§ cubed is eight and the argument of π§ cubed is π.

Finally, we need to determine the value of π§ cubed. Thereβs many different ways of doing this. Since we know the modulus and the argument of π§ cubed, we can just substitute these into the polar form of π§ cubed. This gives us that π§ cubed is equal to eight multiplied by the cos of π plus π sin of π. And we can evaluate this. The cos of π is equal to negative one and the sin of π is equal to zero, which simplifies to give us the value of negative eight. And itβs also worth noting we could have found this value by sketching π§ cubed onto an Argand diagram.

Since the argument of π§ is π, the angle it makes with the positive real axis is π; itβs half a turn. This means it lies on the negative real axis. And since this modulus is equal to eight, itβs at distance eight from the origin. This means itβs equal to negative eight. Therefore, we were able to show if π§ is the complex number one plus root three π, the modulus of π§ is two, the argument of π§ is π by three and the modulus of π§ cubed is eight, the argument of π§ cubed is π, and π§ cubed is just equal to negative eight.