2.9. THE FUNDAMENTAL THEOREM OF ALGEBRA 27

•0

Ar r large• a0

Ar

r small

Thus it is reasonable to believe that for some r dur-ing this shrinking process, the set Ar must hit 0. Itfollows that p(z) = 0 for some z.

For example, consider the polynomial x3 + x+1+ i. It has no real zeros. However, you could let z = r (cos t + isin t) and insert this intothe polynomial. Thus you would want to find a point where

(r (cos t + isin t))3 + r (cos t + isin t)+1+ i = 0+0i

Expanding this expression on the left to write it in terms of real and imaginary parts, youget on the left

r3 cos3 t−3r3 cos t sin2 t + r cos t +1+ i(3r3 cos2 t sin t− r3 sin3 t + r sin t +1

)Thus you need to have both the real and imaginary parts equal to 0. In other words, youneed to have(

r3 cos3 t−3r3 cos t sin2 t + r cos t +1,3r3 cos2 t sin t− r3 sin3 t + r sin t +1)= (0,0)

for some value of r and t. First here is a graph of this parametric function of t for t ∈ [0,2π]on the left, when r = 4. Note how the graph misses the origin 0+ i0. In fact, the closedcurve surrounds a small circle which has the point 0+ i0 on its inside.

-50 0 50

x

-50

0

50

y

-2 0 2

x

-2

0

2

y

-4 -2 0 2 4 6

x

-2

0

2

4

y

r too big r too small r just right

Next is the graph when r = .5. Note how the closed curve is included in a circle whichhas 0+ i0 on its outside. As you shrink r you get closed curves. At first, these closedcurves enclose 0+ i0 and later, they exclude 0+ i0. Thus one of them should pass throughthis point. In fact, consider the curve which results when r = 1.386 which is the graph onthe right. Note how for this value of r the curve passes through the point 0+ i0. Thus forsome t, 1.3862(cos t + isin t) is a solution of the equation p(z) = 0.

Now here is a rigorous proof for those who have studied analysis.Proof. Suppose the nonconstant polynomial p(z) = a0 + a1z+ · · ·+ anzn,an ̸= 0, has

no zero in C. Since lim|z|→∞ |p(z)|= ∞, there is a z0 with

|p(z0)|= minz∈C|p(z)|> 0

Then let q(z) = p(z+z0)p(z0)

. This is also a polynomial which has no zeros and the minimum of

|q(z)| is 1 and occurs at z = 0. Since q(0) = 1, it follows q(z) = 1+akzk +r (z) where r (z)consists of higher order terms. Here ak is the first coefficient which is nonzero. Choose asequence, zn→ 0, such that akzk

n < 0. For example, let −akzkn = (1/n). Then

|q(zn)|=∣∣∣1+akzk + r (z)

∣∣∣≤ 1−1/n+ |r (zn)|= 1+akzkn + |r (zn)|< 1

for all n large enough because |r (zn)| is small compared with∣∣akzk

n∣∣ since |r (zn)| involves

only higher order terms and akzkn < 0. This is a contradiction. ■