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 𝑎 plus 𝑏𝑖. 𝑎 and 𝑏 are real numbers. And we say that 𝑎 is the real part
of the complex number, whereas 𝑏 is the imaginary part. And we know that the polar —
sometimes called trigonometric form of a complex number — is 𝑟 cos 𝜃 plus 𝑖 sin
𝜃. 𝑟 is the modulus and 𝜃 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 𝑖𝜃 is
equal to cos 𝜃 plus 𝑖 sin 𝜃. Now, let’s compare this to the
polar form of a complex number. We can see that if we multiply
through by 𝑟, we get 𝑟𝑒 to the 𝑖𝜃 equals 𝑟 cos 𝜃 plus 𝑖 sin 𝜃. And so, we can write our complex
number 𝑧 as 𝑟𝑒 to the 𝑖𝜃, where 𝑟 is still the modulus and 𝜃 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
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 𝑟𝑒 to the 𝑖𝜃, where 𝑟 is the modulus and 𝜃 is the argument
in radians. The modulus is fairly
straightforward to calculate. For a complex number of the form 𝑎
plus 𝑏𝑖, its modulus is the square root of the sum of the square of 𝑎 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 𝑎 or arctan of the imaginary part divided by the real
In this example, that’s arctan of
negative five root six over two divided by five root two over two, which is negative
𝜋 by three. So the argument for our complex
number is negative 𝜋 by three. We calculated the modulus of 𝑧 to
𝑏 five root two and its argument to be negative 𝜋 by three. So in exponential form, we can say
it’s five root two 𝑒 to the negative 𝜋 by three 𝑖. And at this point, it’s worth
recalling that the argument is periodic with a period of two 𝜋. So we can add or subtract multiples
of two 𝜋 to our argument.
If we add two 𝜋 to negative 𝜋 by
three, we get five root two 𝑒 to the five 𝜋 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 and less than or equal to 𝜋, 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 𝑟𝑒 to the 𝑖𝜃 is the same
as 𝑟 cos 𝜃 plus 𝑖 sin 𝜃, we can distribute these parentheses and then compare
this directly to the algebraic form of a complex number. The real part will be 𝑟 cos 𝜃 and
the imaginary part will be 𝑖 sin 𝜃.
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 𝑟 one 𝑒 to the 𝑖𝜃 one and 𝑟 two 𝑒 to the 𝑖𝜃 two. Their product is 𝑟 one 𝑒 to the
𝑖𝜃 one multiplied by 𝑟 two 𝑒 to the 𝑖𝜃 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 𝑟 one 𝑟 two 𝑒 to the 𝑖 𝜃 one plus 𝜃 two. Essentially, we multiply their
moduli and add their arguments. Similarly, to divide two complex
numbers, we get 𝑟 one divided by 𝑟 two multiplied by 𝑒 to the 𝑖 𝜃 one minus 𝜃
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 by two 𝑖 and 𝑧 two equals six 𝑒 to the 𝜋 by three 𝑖, express 𝑧 one
𝑧 two in the form 𝑎 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 by two. The modulus of our second complex
number is six and its argument is 𝜋 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 by two plus 𝜋 by three.
We can add these two fractions by
creating a common denominator. That’s six. And we get negative three 𝜋 by six
plus two 𝜋 by six which is negative 𝜋 by six. And therefore, we see that 𝑧 one
𝑧 two is 30𝑒 to the negative 𝜋 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 by six plus
𝑖 sin of negative 𝜋 by six. We’ll distribute these
And we see that this is equivalent
to 30 cos of negative 𝜋 by six plus 30 sin of negative 𝜋 by six 𝑖. Now, these are standard
results. Cos of negative 𝜋 by six is root
three over two and sin of negative 𝜋 by six is negative one-half. And we can therefore see that 𝑧
one 𝑧 two — the product of these two complex numbers — 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 𝑎 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 by two radians. And in exponential form, we can say
that this is the same as root two 𝑒 to the 𝜋 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 𝑎, 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 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 by two 𝑖 over root two 𝑒 to the negative 𝜋 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 by two minus negative 𝜋 by four is three 𝜋 by four. In exponential form then, 𝑧 is
equal to 𝑒 to the three 𝜋 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 𝑎𝑒 to the 𝑖𝜃 plus
𝑏𝑒 to the negative two 𝑖𝜃 is equal to cos of two 𝜃 minus five 𝑖 sin of two 𝜃,
where 𝑎 is a real number and 𝑏 is a real number, find 𝑎 and 𝑏.
Here, we have an equation formed of
complex numbers, for which we have some unknowns. Before we can solve for 𝑎 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 𝑎 and 𝑏,
respectively. And their arguments are two 𝜃 and
negative two 𝜃. So we can say that their sum is
equal to 𝑎 cos two 𝜃 plus 𝑖 sin two 𝜃 plus 𝑏 cos of negative two 𝜃 plus 𝑖 sin
of negative two 𝜃.
Now, we’re going to use the fact
that cos 𝜃 is an even function and sin 𝜃 is an odd function. And this means that cos of negative
two 𝜃 is the same as cos of two 𝜃. But sin of negative two 𝜃 is the
same as negative sin of two 𝜃. And we can rewrite our equation
ever so slightly as shown. We need to distribute 𝑎 and 𝑏
over their respective parentheses. And then, we collect like
terms. And we see that we get cos of two
𝜃 times 𝑎 plus 𝑏 plus 𝑖 sin two 𝜃 of 𝑎 minus 𝑏. And of course, comparing this to
original equation, we see that this is equal to cos of two 𝜃 minus five 𝑖 sin two
𝜃. And now, we can equate
Equating coefficients for cos of
two 𝜃, we get one equals 𝑎 plus 𝑏. And for sin of two 𝜃, we get
negative five equals 𝑎 minus 𝑏. We now have a pair of simultaneous
equations in 𝑎 and 𝑏. Let’s add these to eliminate
𝑏. And when we do, we get negative
four equals two 𝑎. So 𝑎 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, 𝑎 is equal to negative
two and 𝑏 is equal to three.
We can of course check this by
popping 𝑎 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 by six 𝑖 plus 𝑒 to the negative 11𝜋 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 𝑟𝑒
to the 𝑖𝜃, its conjugate denoted 𝑧 star is 𝑟𝑒 to the negative 𝑖𝜃. 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 by six 𝑖 and 𝑒 to the negative 11𝜋 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 by six 𝑖 is the complex conjugate for 𝑒 to the 11𝜋 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 𝑟 cos 𝜃. So for our complex number, the real
part is one times cos of 11𝜋 by six. And this means the sum of 𝑒 to the
11𝜋 by six 𝑖 and its conjugate 𝑒 to the negative 11𝜋 by six 𝑖 is two lots of
this. Cos of 11𝜋 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 by six 𝑖 plus 𝑒 to the negative 11𝜋 by six 𝑖 is just
In this video, we’ve learned that
we can express a complex number in exponential form by using 𝑟𝑒 to the 𝑖𝜃, where
𝑟 is the modulus and 𝜃 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