RECITATION 4: Complex Numbers and Euler’s Formula
Goals: In the lecture for unit 4, we discussed a surprising formula, called Euler’s Formula, which relates
the exponential function e x to the trigonometric functions cos x and sin x:
e iθ = cos θ + i sin θ
This formula uses the imaginary constant i, which has the property that i 2 = −1. In this recitation,
we’ll use what we know about Taylor Polynomials (from Recitation 2) to establish this formula as
an identity.
Euler’s Formula
During Recitation 2, you found Taylor polynomials for e x , cos x, and sin x. Those polynomials should
be in your notes. Write the Taylor polynomials here up to the x 5 term:
• ex = 1 + x +
x2 x3 x4 x5
+ +
+
⋯
2
6 24 120
(*)
• cos x =
• sin x =
Group Exercise:
G1. By replacing x with iθ in the starred equation, write a Taylor polynomial for e iθ . Simplify powers
of i whenever possible:
G2. Now write a Taylor polynomial for cos θ + i sin θ, by replacing the x’es with θ’s in the second two
equations, multiplying the equation for sin θ by i, and adding the results together.
G3. Do your results show that e iθ = cos θ + i sin θ?
The Polar Form of a complex number
Every complex number has a Cartesian form and a polar form. The Cartesian form a + bi is the form
you are probably most familiar with, but we can convert this form to polar form re iθ using the same
formulas we saw in Recitation 3:
√
a2 + b2
r =
b
tan θ =
a
Example: Rewrite 1 +
√
3i in polar form.
Solution: Using the conversion formulas, r =
θ = π3 .
√
π
Therefore, 1 + 3i = 2e i 3 .
√
√
√
12 + ( 3)2 = 4 = 2, and tan θ =
√
1
3
, so
To convert back from polar form to Cartesian form, we’d use the formulas a = r cos θ and b = r sin θ.
Polar form is especially useful for raising a complex number to a power.
Example: Find (1 +
√
3i)5 .
Solution 1: We could expand this out:
√
√
√
√
√
√
(1 + 3i)5 = (1 + 3i)(1 + 3i)(1 + 3i)(1 + 3i)(1 + 3i)
Then, we could multiply those five terms together:
√
√
√
√
√
√
√
√
√
√
(1+ 3i)(1+ 3i)(1+ 3i)(1+ 3i)(1+ 3i) = 1+5 3i+10( 3i)2 +10( 3i)3 +5( 3i)4 +( 3i)5
Then, we’d have to simplify all those powers of i:
√
√
√
√
√
√
√
√
√
1+5 3i+10( 3i)2 +10( 3i)3 +5( 3i)4 +( 3i)5 = 1+5 3i−30−30( 3i)+45+9( 3i) = 16−16 3i
But this is very tedious work, and presents lots of opportunities for careless mistakes.
Solution 2: We convert the base from cartesian form to polar form:
(1 +
√ 5
π
5
3i) = (2e 3 i )
Then we use the properties of exponents to simplify the expression:
(2e 3 i )
π
5
= 25 (e 3 i )
π
= 32e
5π
3
i
5
Finally, we convert back to Cartesian form:
x = r cos θ = 32 cos 5π3 = 16 √
y = r sin θ = 32 sin 5π3 = −16 3
So (1 +
√ 5
√
3i) = 16 − 16 3i.
Group Exercise:
G4. In the example above, we raised the polar form (2e 3 ) to the fifth power, and got the polar form
5π
32e 3
π
(a) Discuss what happened to the radius when we raised the number to the fifth power.
(b) Discuss what happened to the angle when we raised the number to the fifth power.
These discussions should help you with the next two exercises:
G5. On a graph of the complex plane, plot (and label) the following points:
z1 = 2i
z2 = −2
1
z3 = − i
2
Then, without actually calculating the squares, estimate the location of z1 2 , z2 2 , and z3 2 . Plot and
label these estimates.
G6. Let z4 = 1 + i. On a clean graph of the complex plane, plot and label z4 , z4 2 , and z4 3 .
The Conjugate and Reciprocals of Complex Numbers
We’ve just seen what happens—both algebraically and graphically—when we raise a complex number
to a positive power. We can use the same procedure to raise a complex number to a negative power,
but if we want to raise it to the negative-one power (that is, find the reciprocal), there’s an even easier
procedure for that:
Example: Let z = 3 + 2i. Find the value of z1 :
Solution 1: We could use the method outlined in the previous section: Convert z into polar
form, raise it to the negative-one power, and then convert back. In order to do this, however,
we’d need a calculator (to figure out tan−1 23 ).
Here’s a solution that doesn’t require a calculator:
Solution 2: The conjugate of a complex number is the number you obtain by changing the
sign of the imaginary part only. In this example, the conjugate of z = 3 + 2i is the number
3 − 2i—the “3” didn’t change, but the “2” did. We use a line over a variable or a number to
indicate its conjugate, so we could write z = 3 − 2i.
To simplify the reciprocal of a complex number z, we can multiply the numerator and the
denominator of the reciprocal by z:
1
1
3 − 2i
3 − 2i 3
2
1
)(
)=
=
=(
= − i
z 3 + 2i
3 + 2i 3 − 2i
13
13 13
Group Exercise:
G7. Of the three members in your group, divide up parts (a), (b), and (c) so that each person completes
one problem. (We want to see three different handwritings here!) If your group has only two
members, it’s okay for the two of you to work together on part (c).
(a) (First team-member:) Let z1 = 2 + 2i. Calculate
form.
1
z1 .
Then convert both z1 and
1
z1
to polar
(b) (Second team-member:) Let z2 = − 21 − 2i . Calculate
form.
1
z2 .
Then convert both z2 and
1
z2
to polar
(c) (Third team-member:) Let z3 = −3 + 3i. Calculate
form.
1
z3 .
Then convert both z3 and
1
z3
to polar
When you have finished your individual tasks, reconvene and complete the next two problems.
G8. Draw the complex plane and the unit circle.
(z1 , z11 , z2 , z12 , z3 , z13 ).
Plot and label the six points you’ve found
G9. Determine a geometric relationship between a complex number z and its reciprocal z1 . (Can you
say anything about their radii? What about their angles?)
Tip: Being able to work with the polar form of a complex number is a skill you’ll use again in Differential
Equations, when determining the amplitude and phase-shift of a solution to a second-order equation.