Math 125,
Fall 2005
N. Kerzman
COMPLEX NUMBERS. ROOTS AND LOGARITHMS.
A ROAD MAP INVOLVING MATLAB.

We assume some familiarity with the most elementary
properties of complex numbers.

Reference.
Analysis.
1.
Saff and Snider, Fundamentals of Complex
A good undergraduate textbook.
THE BASICS .
z=x+iy is a complex number, e.g., z=1+i.
In MatLab you enter z just like that:
>> z=1+i
z =
1.0000 + 1.0000i
>> z=1+2i
z =
1.0000 + 2.0000i
The geometric picture is as follows:
The modulus of z is
z   x 2  y 2 , e.g., 1  i  2 .
In MatLab
notation, abs(z) returns z , and angle(z) returns the polar angle

of z
in radians.
E.g.,
abs(1+2i)
ans =
2.2361
>> angle(1+2i)
1
ans =
1.1071
>> angle (1+i)
ans =
0.7854
Now look at the triangle (0,0), (x,0), (x,y) and recall the
definitions of sin  and cos  from trigonometry. You get x  z cos 
and
y  z sin  .
Hence
z  z (cos   i sin  )
(1.1)
Euler’s Formula
e i  cos   i sin 
(1.2)
together with (1.1), gives the “polar form” of z
z  z ei
(1.3)
You could also write (1.3) in the form
(1.3)’
or even
z  z ei ( 2 )
(1.3)”
z  z ei ( 2k )
k  0,  1,  2,  3,...
because  and   2k have the same sine and cosine.
Where does Euler’s Formula come from? It is motivated by
the familiar power series expansion of e t where instead of t you
write i . Collecting terms and recalling the Taylor series of
cos and sin  , you get (1.2). By power series or by
trigonometric identities, you see that the following familiar
properties of exponentials still hold:
(1.4)
e i (1 2 )  e i1 e i2
(1.5)
e a ib  e a e ib
2
Notice that for any polar angle  ,
(1.6)
e i  1
To see why, use
Euler’s Formula
Examples of this very useful
and
sin 2   cos 2   1
“polar form”.
1  i  2 e i / 4
1
1 i / 4

e
1 i
2
(1  i ) 3  ( 2 ) 3 e i ( / 4 ) 3




z 5  z e i 5 , and
5
if z is expressed as in (1.3), then, e.g.,
in general, for any integer N
z N  z e iN
N
(1.7)
Make sure you verify and understand the formulas above, They are
elementary but revealing and deserve some focused attention.
(1.7) shows that WHEN YOU RAISE z TO THE N-TH POWER, THE MODULUS
GETS RAISED TO THE N-POWER BUT THE POLAR ANGLE GETS MULTIPLIED BY
N.
2.
N-TH ROOTS OF Z.
2.1
N-th roots of 1
Who wants N 1 ? Obviously N 1 =1. Yes, but we want to find all
possible such roots for reasons we’ll see in a moment. Let w be
one of the sought roots. In polar form, w  w ei and 1  1e i 0 .
Taking power N, by (1.7),
necessarily, w  1 .
w  1.
w N  w e iN  1N e iN .
N
Hence,
An obvious  is   0 . We get the root
Let’s call it w0 .
i
Another possibility is  
2
i
2
2
which
N
N
gives a new root w1  e N . Checking: ( w1) N  e N  1 and it
works. We used (1.7). By the same token, we also get another
i
2
2
i
2
2N
candidate w2  e N . Checking: ( w2) N  e N
 e i 4  1 . It checks,
too. We can continue. For any k  0,1,2,3,... a root of 1 appears
and is denoted by
wk  e
i
2
k
N
.
3
Example for N=6.
w0  e
w1  e
i
2
0
6
2
i 1
6
w2  e
w3  e
w4  e
w5  e
2
i 2
6
2
i 3
6
2
i 4
6
2
i 5
6
=
1
=
0.5000 + 0.8660i
=
-0.5000 + 0.8660i
=
-1.0000 + 0.0000i
=
-0.5000 - 0.8660i
=
0.5000 - 0.8660i
There is no need to continue. We would get a new cyle: w6=w0,
w7=w1, w8=w2, etc. If you want to understand why, focus your
attention on the exponents you get:
w0  e
w1  e
i
2
0
6
2
i 1
6
w2  e
2
i 2
6
=
=
=
w6  e
2
( 0 6 )
6
i
w7  e
w8  e
2
i (1 6 )
6
2
i ( 26 )
6
= 1
=0.5000 + 0.8660i
=-0.5000 + 0.8660i
etc.
NOTICE:
OF w1 :
YOU CAN EXPRESS ALL SIX ROOTS IN TERMS OF POWERS
w0  ( w1) 0 , w1  ( w1)1 , w2  ( w1) 2 , w3  ( w1) 3 , w4  ( w1) 4 , w5  ( w1)5
Are there additional
6
1 ?
NO
Reason: If w 6  1 then w is a solution of the polynomial
w 6  1  0 . This polynomial has degree 6. Algebra says
equation
that if the polynomial’s degree is N, you cannot have more than N
solutions.
4
A nice picture. The six sixth roots roots of 1 we got: w0,
w1,…,w5 are the vertices of a regular polygon (hexagon) of
center 0, inscribed in the circle of radius 1.
A totally analogous situation holds for the N Nth roots of
1. Everything said is valid, and for the same reasons, once
you replace 6 by N.
The roots are w0,w1,w2,…,w(N-1); w1
is very useful:
Each root is obtained by taking the
corresponding power of w1.
2.2
Nth roots of an arbitrary z=x+iy
When z=0, the only possible root is w=0. We consider z  0 .
Express z in polar form z  z ei . One particular Nth root of
z, let’s call it R, comes to mind right away:
i
Checking: R

RN z eN
(2.1)
N
( z ) e
N
Example: for z=1+i
N
i

N
N
 z ei  z
z  2 and  
In MatLab notation,
>> z=1+i;
>> mod=abs(z);
>> theta=angle(z);
>> R=(mod^(1/6))*exp(i*theta/6);
>> R
R =
1.0504 + 0.1383i
>> %checking:
>> R^6
R^6
ans =
1.0000 + 1.0000i
5

4
.
IT IS NOW VERY EASY TO FIND THE OTHER 6TH ROOTS OF z=1+i
USING THE 6th ROOTS OF 1 DISCUSSED IN SECTION (2.1)
You just make the products
R*w0, R*w1, R*w2, R*w3,R*w4, R*w5.
Each is a 6th root of z=1+i. Reason: for, e.g., R*w3:
(R*w3)^6=(R^6)*(w3^6)=z*1=z. That’s it!
Notice:
a) R*w0=R is the first root we got. It is also the one MatLab
returns when you enter z^(1/6)
b) all six roots listed are different because the factor R is not
zero.
c) the list is complete. There are no other 6th roots of z=1+i.
The reason is very nice and simple. Can you see it?
Summing up:
 To find the Nth roots of a given z=x+iy, find first R via
formula (2.1). This step involves knowing z, of course.
Next mutiply R by each Nth root of 1 (these roots of 1 have
nothing to do with z). The N products you get
R*w0,
R*w1, R*w2, …,R*w(N-1) are what you want.

MatLab returns R*w0 when you enter z^(1/N)
The picture you get for the Nth roots of z: the roots are the
vertices of a regular polygon inscribed in the circle of center 0
and radius
N
z .
The polygon is rotated with respect to the one
for the Nth roots of 1 by an angle  angle ( z ) / N
3. THE LOGARITHMS OF z=x+iy
The logarithms under discussion are to base e=2.71…
w=log(z) means: w is a complex number such that e w  z
w
u iv
 e u e iv . A natural candidate to be log(z)
If w= u+iv, then e  e
appears (provided that z  0 ). Namely, for z  z ei
L= log( z )  i
(3.1)
Checking:
eL  e
e  z ei  z
log( z ) i
6
z=0 is excluded because you cannot define log(0).
L is the logarithm that MatLab returns when you enter log(z)
Example
log(1+2i)
ans =
0.8047 + 1.1071i
Let’s compare with the discussion above and see that MatLab gave
us precisely our L.
>> z=1+2i
z =
1.0000 + 2.0000i
>> u=log(abs(z))
u =
0.8047
>> v=angle(z)
v =
1.1071
>> L=u+i*v
L =
0.8047 + 1.1071i
>> %checking
>> exp(L)
ans =
1.0000 + 2.0000i
z=x+iy HAS MANY (INFINITELY MANY) LOGARITHMS
7
L+2*pi*k*i is also a logarithm of z for any k  0,  1,  2,  3,... .
Let’s call it Lk and check:
exp(Lk)=exp(L)*exp(2*pi*k*i)=z.
exp(2*pi*k*i)=1.
We used that
exp(L)=z and
All this is analog to what we did with roots, even a bit easier.
L plays a role similar to that of R in formula (2.1). The Lk are
the analogs of w0,w1,w2, etc. Differences are
a)the Lk are infinitely many, all different; for each k there is
one Lk, Since there are infinitely many k, there are infinitely
many Lk.
b) the Lk are not the vertices of a regular polygon. They all
lie on the same vertical line, a line parallel to the vertical
coordinate axis at a distance log(abs(z)) from it.
Summing up.

Any z=x+iy other than z=0 has infinitely many logarithms,
namely: Lk= log(abs(z))+angle(z)*i + 2*pi*k*i, one for
each k  0,  1,  2,  3,...

MatLab returns L=log(abs(z))+i*angle(z)
log(z).

z=0 has no log(z) because there is no w such that exp(w)=0.
when you enter
There are no other logarithms of z. I.e., no w other than the
listed ones satisfies exp(w)=z. This is not too difficult to
see. The mathematically inclined may like to try to figure
this one out by themselves.
8