Video: Exponential Form of a Complex Number

In this video, we will learn how to convert a complex number from the algebraic to the exponential form (Euler’s form) and vice versa.

14:59

Video Transcript

In this video, we’ll learn how to express complex numbers in exponential form. We should know how to express a complex number in algebraic and polar form. So this is a natural extension of this logic. We’ll learn what we mean by exponential form and how to multiply and divide with these numbers. We’ll also learn how to convert between numbers in algebraic, polar, and exponential form before discovering how the exponential form can help us solve equations involving complex numbers.

The algebraic form of a complex number is 𝑧 equals π‘Ž plus 𝑏𝑖. π‘Ž and 𝑏 are real numbers. And we say that π‘Ž is the real part of the complex number, whereas 𝑏 is the imaginary part. And we know that the polar β€” sometimes called trigonometric form of a complex number β€” is π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. π‘Ÿ is the modulus and πœƒ is the argument usually given in radians. So what about the exponential form of a complex number?

Here, we need Euler’s formula. This says that 𝑒 to the π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. Now, let’s compare this to the polar form of a complex number. We can see that if we multiply through by π‘Ÿ, we get π‘Ÿπ‘’ to the π‘–πœƒ equals π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. And so, we can write our complex number 𝑧 as π‘Ÿπ‘’ to the π‘–πœƒ, where π‘Ÿ is still the modulus and πœƒ is still the argument, given in radians here. And we can use the same methods for calculating the modulus and argument of a complex number in exponential form as we would for a complex number written in polar form. Let’s see what this might look like.

Put the number 𝑧 equals five root two over two minus five root six over two 𝑖 in exponential form.

This complex number is currently in algebraic form. It has a real part of five root two over two and an imaginary part of negative five root six over two. Remember a complex number in exponential form is π‘Ÿπ‘’ to the π‘–πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument in radians. The modulus is fairly straightforward to calculate. For a complex number of the form π‘Ž plus 𝑏𝑖, its modulus is the square root of the sum of the square of π‘Ž and 𝑏.

In this case, it’s the square root of five root two over two all squared plus negative five root six over two all squared. Five root two over two all squared is 25 over two. And negative five root six over two all squared is 75 over two. The sum of 25 over two and 75 over two is 100 over two, which is simply 50. So the modulus of 𝑧 is the square root of 50, which we can simplify to five root two. But what about the argument?

If we put this complex number on the Argand plane, it’s represented by the point whose Cartesian coordinates are five root two over two and negative five root six over two. This means it lies in the fourth quadrant. We can find the argument for complex numbers that lie in the first and fourth quadrant by using the formula arctan of 𝑏 divided by π‘Ž or arctan of the imaginary part divided by the real part.

In this example, that’s arctan of negative five root six over two divided by five root two over two, which is negative πœ‹ by three. So the argument for our complex number is negative πœ‹ by three. We calculated the modulus of 𝑧 to 𝑏 five root two and its argument to be negative πœ‹ by three. So in exponential form, we can say it’s five root two 𝑒 to the negative πœ‹ by three 𝑖. And at this point, it’s worth recalling that the argument is periodic with a period of two πœ‹. So we can add or subtract multiples of two πœ‹ to our argument.

If we add two πœ‹ to negative πœ‹ by three, we get five root two 𝑒 to the five πœ‹ over three 𝑖. Whilst the argument of the complex number in this second form is outside of the range for the principal argument which is greater than negative πœ‹ and less than or equal to πœ‹, it’s not unusual to see these numbers written in either form. And what about converting back from a number in exponential form?

Well, the conversion between exponential and polar form is fairly straightforward, which is to use the same values for the modulus and the argument. To convert from exponential form back into algebraic form though, we first convert into polar form and then convert it to algebraic form. Since π‘Ÿπ‘’ to the π‘–πœƒ is the same as π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, we can distribute these parentheses and then compare this directly to the algebraic form of a complex number. The real part will be π‘Ÿ cos πœƒ and the imaginary part will be 𝑖 sin πœƒ.

Now that we have a definition of the exponential form for a complex number, we can use this to develop some rules for multiplication and division with these numbers. Let’s say we have two complex numbers π‘Ÿ one 𝑒 to the π‘–πœƒ one and π‘Ÿ two 𝑒 to the π‘–πœƒ two. Their product is π‘Ÿ one 𝑒 to the π‘–πœƒ one multiplied by π‘Ÿ two 𝑒 to the π‘–πœƒ two. And then, we recall the properties of the modulus and arguments of a complex number.

The modulus of the product of two complex numbers is equal to the product of their moduli and the argument of their product is equal to the sum of their respective arguments. So we can say that the product of 𝑧 one and 𝑧 two is π‘Ÿ one π‘Ÿ two 𝑒 to the 𝑖 πœƒ one plus πœƒ two. Essentially, we multiply their moduli and add their arguments. Similarly, to divide two complex numbers, we get π‘Ÿ one divided by π‘Ÿ two multiplied by 𝑒 to the 𝑖 πœƒ one minus πœƒ two. This time, we divide their moduli and subtract their arguments.

Now whilst it does look like we could have simply applied the rules for integer exponents to derive these results, we do need to be a little careful assuming those rules work for all complex numbers. This is not always necessarily true. And so, it’s much more preferable to think about the product and quotient of complex numbers in terms of their moduli and arguments. Let’s look at how to apply these processes to multiplication and division of complex numbers in exponential form.

Given 𝑧 one equals five 𝑒 to the negative πœ‹ by two 𝑖 and 𝑧 two equals six 𝑒 to the πœ‹ by three 𝑖, express 𝑧 one 𝑧 two in the form π‘Ž plus 𝑏𝑖.

Here, we’ve been given two complex numbers in exponential form and we’re being asked to find their product in algebraic form. It’s much simpler to multiply complex numbers whilst they are in exponential form. So we’ll do that bit first before converting their product to algebraic form. To multiply two complex numbers in exponential form, we multiply their moduli and add their arguments.

The modulus of our first complex number is five and its argument is negative πœ‹ by two. The modulus of our second complex number is six and its argument is πœ‹ by three. This means the modulus of the product of these two complex numbers will be five times six which is 30. And the argument of 𝑧 one 𝑧 two will be negative πœ‹ by two plus πœ‹ by three.

We can add these two fractions by creating a common denominator. That’s six. And we get negative three πœ‹ by six plus two πœ‹ by six which is negative πœ‹ by six. And therefore, we see that 𝑧 one 𝑧 two is 30𝑒 to the negative πœ‹ by six 𝑖. And that’s in exponential form. So how do we convert this into algebraic form? The easiest way is to represent it in polar form first. It’s 30 cos negative πœ‹ by six plus 𝑖 sin of negative πœ‹ by six. We’ll distribute these parentheses.

And we see that this is equivalent to 30 cos of negative πœ‹ by six plus 30 sin of negative πœ‹ by six 𝑖. Now, these are standard results. Cos of negative πœ‹ by six is root three over two and sin of negative πœ‹ by six is negative one-half. And we can therefore see that 𝑧 one 𝑧 two β€” the product of these two complex numbers β€” simplifies to 15 root three minus 15𝑖. This is now in algebraic form as required. If we compare it to the general form in our question, we see that π‘Ž is 15 root three and 𝑏 is negative 15.

Given that 𝑧 equals 𝑖 root two over one minus 𝑖, write 𝑧 in exponential form.

To answer this question, we have two options. We could divide these two complex numbers in algebraic form. And to do this, we need to multiply both the numerator and denominator by the conjugate of the denominator then distribute and simplify as far as possible. I’m sure you’ll agree that’s rather a lengthy process. Instead, we’ll choose to write these complex numbers in exponential form. So we’ll need to calculate their moduli and arguments.

𝑖 root two is a purely imaginary number. On an Argand diagram, it’s represented by the point whose Cartesian coordinates are zero, root two. Its modulus is the length of the line segment that joins this point to the origin. So it’s root two. And since the argument is measured in the counterclockwise direction from the positive real axis, we can see that the argument of this complex number is equivalent to 90 degrees. That’s πœ‹ by two radians. And in exponential form, we can say that this is the same as root two 𝑒 to the πœ‹ by two 𝑖.

The complex number one minus 𝑖 is a little more tricky. Its real part is positive and its imaginary part is negative. So it lies in the fourth quadrant. Now, its modulus is independent of this fact. We simply use the formula the square root of the sum of the square of the real and imaginary parts. So that’s the square root of one squared plus negative one squared which once again is the square root of two.

We do need to be a little bit more careful with the argument. Since it’s in the fourth quadrant, we can use the formula that’s unique to complex numbers that are plotted in the first and fourth quadrants. That’s arctan of 𝑏 over π‘Ž, arctan of the imaginary part divided by the real part. So in this case, that’s the arctan of negative one over one which is negative πœ‹ by four. We expected a negative value for the argument as this time we’re measuring in a clockwise direction. And so, we can rewrite our fraction as root two 𝑒 to the πœ‹ by two 𝑖 over root two 𝑒 to the negative πœ‹ by four 𝑖. And now, we can easily divide.

To divide complex numbers in exponential form, we divide their moduli and subtract their arguments. Root two divided by root two is one and πœ‹ by two minus negative πœ‹ by four is three πœ‹ by four. In exponential form then, 𝑧 is equal to 𝑒 to the three πœ‹ by four 𝑖. We’ve seen how to multiply and divide with complex numbers in exponential form. Let’s now look at how to use properties of complex numbers in exponential form to solve equations.

Given that π‘Žπ‘’ to the π‘–πœƒ plus 𝑏𝑒 to the negative two π‘–πœƒ is equal to cos of two πœƒ minus five 𝑖 sin of two πœƒ, where π‘Ž is a real number and 𝑏 is a real number, find π‘Ž and 𝑏.

Here, we have an equation formed of complex numbers, for which we have some unknowns. Before we can solve for π‘Ž and 𝑏, we’ll need to ensure that each complex number is in the same form. Let’s convert the left-hand side to polar form. It’s made up of two complex numbers. Their moduli are π‘Ž and 𝑏, respectively. And their arguments are two πœƒ and negative two πœƒ. So we can say that their sum is equal to π‘Ž cos two πœƒ plus 𝑖 sin two πœƒ plus 𝑏 cos of negative two πœƒ plus 𝑖 sin of negative two πœƒ.

Now, we’re going to use the fact that cos πœƒ is an even function and sin πœƒ is an odd function. And this means that cos of negative two πœƒ is the same as cos of two πœƒ. But sin of negative two πœƒ is the same as negative sin of two πœƒ. And we can rewrite our equation ever so slightly as shown. We need to distribute π‘Ž and 𝑏 over their respective parentheses. And then, we collect like terms. And we see that we get cos of two πœƒ times π‘Ž plus 𝑏 plus 𝑖 sin two πœƒ of π‘Ž minus 𝑏. And of course, comparing this to original equation, we see that this is equal to cos of two πœƒ minus five 𝑖 sin two πœƒ. And now, we can equate coefficients.

Equating coefficients for cos of two πœƒ, we get one equals π‘Ž plus 𝑏. And for sin of two πœƒ, we get negative five equals π‘Ž minus 𝑏. We now have a pair of simultaneous equations in π‘Ž and 𝑏. Let’s add these to eliminate 𝑏. And when we do, we get negative four equals two π‘Ž. So π‘Ž must be equal to negative two. And then, we substitute this back into the first equation. And we get one equals negative two plus 𝑏. So 𝑏 must be equal to three. Therefore, π‘Ž is equal to negative two and 𝑏 is equal to three.

We can of course check this by popping π‘Ž equals negative two and 𝑏 equals three into the other equation. When we do, we see that negative two minus three equals negative five as required. In our final example, we’re going to recall the properties of the complex conjugate and see how being able to spot the complex conjugate for numbers in exponential form can save us some time.

Find the numerical value of 𝑒 to the 11πœ‹ by six 𝑖 plus 𝑒 to the negative 11πœ‹ by six 𝑖.

To evaluate the sum of these two complex numbers, we could convert them into algebraic form and add simply by collecting like terms. However, it is useful to be able to spot the complex conjugate of a number written in exponential form and we’ll see why in a moment. For a complex number 𝑧 equals π‘Ÿπ‘’ to the π‘–πœƒ, its conjugate denoted 𝑧 star is π‘Ÿπ‘’ to the negative π‘–πœƒ. Notice how the modulus of the conjugate is the same as the modulus of the original complex number and that its argument is the negative of the argument of the original complex number.

The two complex numbers we have 𝑒 to the 11πœ‹ by six 𝑖 and 𝑒 to the negative 11πœ‹ by six 𝑖 both have a modulus of one. But the argument of the second complex number is the negative argument of the first and in fact vice versa. This means that 𝑒 to the negative 11πœ‹ by six 𝑖 is the complex conjugate for 𝑒 to the 11πœ‹ by six 𝑖 and vice versa. But why does this help? Well, it allows us to use the rule for addition of a complex number and its conjugate. The sum of a complex number and its conjugate is equal to two times the real part of that complex number.

Now, the real part of a complex number written in exponential or polar form is simply π‘Ÿ cos πœƒ. So for our complex number, the real part is one times cos of 11πœ‹ by six. And this means the sum of 𝑒 to the 11πœ‹ by six 𝑖 and its conjugate 𝑒 to the negative 11πœ‹ by six 𝑖 is two lots of this. Cos of 11πœ‹ by six is root three over two. So two times the real part of our complex number is two times root three over two, which is simply root three. And we can see that the numerical value of 𝑒 to the 11πœ‹ by six 𝑖 plus 𝑒 to the negative 11πœ‹ by six 𝑖 is just root three.

In this video, we’ve learned that we can express a complex number in exponential form by using π‘Ÿπ‘’ to the π‘–πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument expressed in radians. And we’ve seen that working with numbers in this form can help us simplify calculations, involving multiplication and division. To multiply two complex numbers, for example, we multiply their moduli and we add their arguments. And to divide two complex numbers written in exponential form, we divide their moduli and subtract their arguments.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.