1.6: Euler's Formula
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Euler’s (pronounced ‘oilers’) formula connects complex exponentials, polar coordinates, and sines and cosines. It turns messy trig identities into tidy rules for exponentials. We will use it a lot. The formula is the following:
eiθ=cos(θ)+isin(θ).
There are many ways to approach Euler’s formula. Our approach is to simply take Equation ??? as the definition of complex exponentials. This is legal, but does not show that it’s a good definition. To do that we need to show the eiθ obeys all the rules we expect of an exponential. To do that we go systematically through the properties of exponentials and check that they hold for complex exponentials.
eiθ behaves like a true exponential
eit differentiates as expected:
deitdt=ieit.
- Proof
-
This follows directly from the definition in Equation ???:
deitdt=ddt(cos(t)+isin(t))=−sin(t)+icos(t)=i(cos(t)+isin(t))=ieit.
ei⋅0=1.
- Proof
-
This follows directly from the definition in Equation ???:
ei⋅0=cos(0)+isin(0)=1.
The usual rules of exponents hold:
eiaeib=ei(a+b).
- Proof
-
This relies on the cosine and sine addition formulas and the definition in Equation ???:
eia⋅eib=(cos(a)+isin(a))⋅(cos(b)+isin(b))=cos(a)cos(b)−sin(a)sin(b)+i(cos(a)sin(b)+sin(a)cos(b))=cos(a+b)+isin(a+b)=ei(a+b).
The definition of eiθ is consistent with the power series for ex.
- Proof
-
To see this we have to recall the power series for ex, cos(x) and sin(x). They are
ex=1+x+x22!+x33!+x44!+...cos(x)=1−x22!+x44!−x66!+…sin(x)=x−x33!+x55!+...
Now we can write the power series for eiθ and then split it into the power series for sine and cosine:
eiθ=∞∑0(iθ)nn!=∞∑0(−1)kθ2k(2k)!+i∞∑0(−1)kθ2k+1(2k+1)!=cos(θ)+isin(θ).
So the Euler formula definition is consistent with the usual power series for ex.
Properties P1-P4 should convince you that eiθ behaves like an exponential.
Complex Exponentials and Polar Form
Now let’s turn to the relation between polar coordinates and complex exponentials.
Suppose z=x+iy has polar coordinates r and θ. That is, we have x=rcos(θ) and y=rsin(θ). Thus, we get the important relationship
z=x+iy=rcos(θ)+irsin(θ)=r(cos(θ)+isin(θ))=reiθ.
This is so important you shouldn’t proceed without understanding. We also record it without the intermediate equation.
z=x+iy=reiθ.
Because r and θ are the polar coordinates of (x,y) we call z=reiθ the polar form of z.
Let’s now verify that magnitude, argument, conjugate, multiplication and division are easy in polar form.
|eiθ|=1.
- Proof
-
|eiθ|=|cos(θ)+isin(θ)|=√cos2(θ)+sin2(θ)=1.
In words, this says that eiθ is always on the unit circle - this is useful to remember!
Likewise, if z=reiθ then |z|=r. You can calculate this, but it should be clear from the definitions: |z| is the distance from z to the origin, which is exactly the same definition as for r.
If z=reiθ then arg(z)=θ.
- Proof
-
This is again the definition: the argument is the polar angle θ.
¯(z=reiθ)=re−iθ.
- Proof
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¯(z=reiθ)=¯r(cos(θ)+isin(θ))=r(cos(θ)−isin(θ))=r(cos(−θ)+isin(−θ))=re−iθ.
In words: complex conjugation changes the sign of the argument.
If z1=r1eiθ1 and z2=r2eiθ2 then
z1z2=r1r2ei(θ1+θ2).
This is what mathematicians call trivial to see, just write the multiplication down. In words, the formula says the for z1z2 the magnitudes multiply and the arguments add.
Again it's trivial that
r1eiθ1r2eiθ2=r1r2ei(θ1−θ2).
Here’s a simple but important example. By looking at the graph we see that the number 2i has magnitude 2 and argument π/2. So in polar coordinates it equals 2eiπ/2. This means that multiplication by 2i multiplies lengths by 2 and add π/2 to arguments, i.e. rotates by 90∘. The effect is shown in the figures below
Let's compute (1+i)6 and (1+i√32)3
Solution
1+i has magnitude = √2 and arg=π/4, so 1+i=√2eiπ/4. Rasing to a power is now easy:
(1+i)6=(√2eiπ/4)6=8e6iπ/4=8e3iπ/2=−8i.
Similarly, 1+i√32=eiπ/3, so (1+i√32)3=(1⋅eiπ/3)3=eiπ=−1
Complexification or Complex Replacement
In the next example we will illustrate the technique of complexification or complex replacement. This can be used to simplify a trigonometric integral. It will come in handy when we need to compute certain integrals.
Use complex replacement to compute
I=∫excos(2x) dx.
Solution
We have Euler's formula
e2ix=cos(2x)+isin(2x),
so cos(2x)=Re(e2ix). The complex replacement trick is to replace cos(2x) by e2ix. We get (justification below)
Ic=∫excos2x+iexsin2x dx
with
I=Re(Ic)
Computing Ic is straightforward:
Ic=∫exei2x dx=∫ex(1+2i) dx=ex(1+2i)1+2i.
Here we will do the computation first in rectangular coordinates. In applications, for example throughout 18.03, polar form is often preferred because it is easier and gives the answer in a more useable form.
Ic=ex(1+2i)1+2i⋅1−2i1−2i=ex(cos(2x)+isin(2x))(1−2i)5=15ex(cos(2x)+2sin(2x)+i(−2cos(2x)+sin(2x)))
So,
I=Re(Ic)=15ex(cos(2x)+2sin(2x)).
Justification of complex replacement. The trick comes by cleverly adding a new integral to I as follows, Let J=∫exsin(2x) dx. Then we let
Ic=I+iJ=∫ex(cos(2x)+isin(2x)) dx=∫ex22ix dx.
Clearly, by construction, Re(Ic)=I as claimed above.
Alternative using polar coordinates to simplify the expression for Ic:
In polar form, we have 1+2i=reiϕ, where r=√5 and ϕ=arg(1+2i)=tan−1(2) in the first quadrant. Then:
Ic=ex(1+2i)√5eiϕ=ex√5ei(2x−ϕ)=ex√5(cos(2x−ϕ)+isin(2x−ϕ)).
Thus,
I=Re(Ic)=ex√5cos(2x−ϕ).
Nth roots
We are going to need to be able to find the nth roots of complex numbers, i.e., solve equations of the form
zN=c,
where c is a given complex number. This can be done most conveniently by expressing c and z in polar form, c=Reiϕ and z=reiθ. Then, upon substituting, we have to solve
rNeiNθ=Reiϕ
For the complex numbers on the left and right to be equal, their magnitudes must be the same and their arguments can only differ by an integer multiple of 2π. This gives
r=R1/N(Nθ=ϕ+2πn),wheren=0,±1,±2,...
Solving for θ, we have
θ=ϕN+2πnN.
Find all 5 fifth roots of 2.
Solution
For c=2, we have R=2 and ϕ=0, so the fifth roots of 2 are
zn=21/5e2nπi/5, where n=0,±1,±2,...
Looking at the right hand side we see that for n=5 we have 21/5e2πi which is exactly the same as the root when n=0, i.e. 21/5e0i. Likewise n=6 gives exactly the same root as n=1, and so on. This means, we have 5 different roots corresponding to n=0,1,2,3,4.
zn=21/5,e1/5e2πi/5,e1/5e4πi/5,e1/5e6πi/5,e1/5e8πi/5
Similarly we can say that in general c=Reiϕ has N distinct N th roots:
zn=r1/Neiϕ/N+i2π(n/N) for n=0,1,2,...,N−1.
Find the 4 forth roots of 1.
Solution
We need to solve z4=1, so ϕ=0. So the 4 distinct fourth roots are in polar form
zn=1,eiπ/2,eiπ,ei3π/2
and in Cartesian representation
zn=1,i,−1,−i.
Find the 3 cube roots of -1.
Solution
z2=−1=eiπ+i2πn. So, zn=eiπ+i2π(n/3) and the 3 cube roots are eiπ/3, eiπ, ei5π/3. Since π/3 radians is 60∘ we can simplify:
eiπ/3=cos(π/3)+isin(π/3)=12+i√32⇒zn=−1,12±i√32
Find the 5 fifth roots of 1+i.
Solution
z5=1+i=√2ei(π/4+2nπ)
for n=0,1,2,.... So, the 5 fifth roots are
21/10eiπ/20, 21/10ei9π/20, 21/10ei17π/20, 21/10ei25π/20, 21/10ei33π/20.
Using a calculator we could write these numerically as a+bi, but there is no easy simplification.
We should check that our technique works as expected for a simple problem. Find the 2 square roots of 4.
Solution
z2=4ei2πn. So, zn=2eiπn, with n=0,1. So the two roots are 2e0=2 and 2eiπ=−2 as expected!
geometry of Nth roots
Looking at the examples above we see that roots are always spaced evenly around a circle centered at the origin. For example, the fifth roots of 1+i are spaced at increments of 2π/5 radians around the circle of radius 21/5.
Note also that the roots of real numbers always come in conjugate pairs.