Gaussian Integers & Division Algorithm Project

By: Clay Kitchings

MATH 6000

(Also EMAT 6690)


Motivation for this write-up comes from my MATH 6000 project/assignment on the Gaussian Integers.  This will become a webpage “essay” for EMAT 6690.  This assignment makes use of the following technologies: TI84 Plus Silver Edition (for “efficient” multiplication of complex numbers), Geometer’s Sketchpad (for plotting points and for other graphics). Note: This link will NOT be activated until after 11:00am, Thursday, May 3rd.  No one (teacher, student, or other individual or group) other than Clay Kitchings will view this document in any shape or form until after the 11:00am deadline. 


A Gaussian Integer is a complex number a + bi where a and b are both integers. 


Let this set be represented by Z[i] = {a+bi |  a, b are in Z}.


Like Z and F[x] (and a few other rings) Z[i] has a division algorithm. Our first goal is to see how the division algorithm works:


The Division Algorithm, also known as “Proposition 3.1” states the following:  “Let z = a + bi and w = c + di (z, w, elements of Z[i]), . Then there are Gaussian integers q and r so that z = qw + r with |r|2 < |w|2 ” (Shifrin, 1996, p. 140).


For our first example, let z = 4 + 7i and w = -1 + 4i.


For the record, we’ll compute some multiples of w. We’ll use the TI84 to expedite the process. L2 = L1*w:







1)   On the following grid, we shall mark all of the points in <w>, the ideal generated by w.  Note that the horizontal axis is the real axis and the vertical axis is the imaginary axis (though it is only labeled as a typical x-y plane).

2)   We’ll connect some of the points in <w> and tile the plane with squares. This process was initiated by drawing the polygon with vertices 0, w, iw, and (1+i)w.

3)   Next, we’ll plot z in the complex plane (as shown below).


4)   The yellow-colored point z now lies in the interior of one of the squares above. Since the vertex is in <w> it is of the form qw for some q in Z[i].  The vertex of the square nearest to z is (1-i), so q = 1-i.


5)   Now let r = z – qw. What is r? 



6)   Claim: We have just applied the division algorithm to z and w.


Discussion: We have applied the division algorithm because we’ve expressed z as a product of w and a Gaussian integer plus some remainder (also a Gaussian integer). All of the vertices with blue points are in <w>. The w vertex itself is also highlighted yellow.



Note (visual observation): we may observe a pattern along each line passing through the points in <w>.  For example, after calculating various multiples of w (i.e., various forms of qw), we find that a pattern exists within the real and imaginary parts of the complex number q. For example, the points which are collinear with w and (1+i)w all have 1 as the real part of q. We’ll call that line L1.  The “parallel” line immediately below L1 has 0 for all the real parts of each complex number q. Immediately below and “parallel” to that line, we find a similar result for the real parts of q, except for now it is -1. As we proceed downward in the plane, we find that for each line the real part of the number q decreases by one. 


There is a similar pattern that occurs for those lines which are perpendicular to L1.  As one moves further to the left in the plane, one finds the imaginary part of the number q increases by one (or rather, the coefficient of i increases by one for the complex number q). Moving to the right causes the coefficient of i to decrease by one with each additional line to the right in the plane. This pattern is useful in more quickly determining the vertices of the squares (all of which are in <w>).




Now repeat the steps above with z = -1 + 4i and w = 1 + 2i. Why are we using these numbers? They come from the above problem’s continuation of the division algorithm.


From above, we have the following:

       r = z – qw       However, if we are to continue to “divide,” we would then repeat the process as follows:


       z = w – q’w’    for some new q’, w’ that are in Z[i].  This process is somewhat analogous to finding the gcd(14, 35) among the integers by using the Euclidean Algorithm. 





The above graphic illustrates the complex number w along with the ideal generated by w (or <w>). The pink-colored point z lies in the interior of one of the squares, and it is closest to the vertex (1+i)w. This vertex is a part of <w>, so it is of the form qw for some q in Z[i].  This means q=1+i.


Let r = z – qw, and solve for r:       r= (-1+4i) – (1+i)(1+2i) Ź    r= i = (0 + 1i)


Calculate gcd(4+7i, -1+4i).


z      =     (q)   (w)   +    r


4+7i = (-1+4i)(1-i) + (1+2i)


-1+4i = (1+2i)(1+i) +   i

1+2i = i(-i+2) + 0


So, gcd(4+7i, -1+4i) = i. 


Now, write the gcd as a linear combination of (4+7i) and (-1+4i).  Let these two complex numbers be z and w as they were earlier in the problem.  So now we need to write i as a linear combination of z and w.


i = w – (1+2i)(1+i)

i = w – [z – w(1-i)] [1+i] = w – [z(1+i) – w(1-i)(1+i)] = w – z(1+i) + w(2) = 3w – z(1+i)


So i = (3+0i)(-1+4i) + (4+7i)(-1-i)




Next, we shall find the gcd (7-4i, 1+3i).


7-4i = (1+3i)( q ) + r


Use a diagram/grid:




Note: z may not have one unique “nearest” vertex, so this number need not be unique.


Take q = -3i. 


7-4i = (1+3i)( q ) + r


7-4i = (1+3i)(-3i) + r Ź r = -2-i.



7-4i = (1+3i)(-3i) + (-2-i)


Now we consider “dividing” (1+3i) and (-2-i):



It appears Z may actually be one of the vertices of a square in the ideal this time. (Note: this z and w are different from the one’s we started with.) The vertex “nearest” z is (-1-i)w, so q=(-1-i). Let’s now divide again:


7-4i = (1+3i)(-3i) + (-2-i)

1+3i = (-2-i)(-1-i) + r’           where r’ is some new remainder

Ź r = 0


Therefore, the gcd(7-4i, 1+3i) = -2-i. Now we shall write it as a linear combination of z and w:


-2-i = z – (-3i)(1+3i)      Ź    -2-i = (7-4i) + (3i)(1+3i)










Shifrin, Theodore. (1996). Abstract algebra: A geometric approach. Upper Saddle River, NJ. Prentice Hall.