1987 AMC 12 第 28 题

Question

设

a

、

b

、

c

、

d

为实数。假设方程

z^4+az^3+bz^2+cz+d=0

的所有根都是复数，并且都位于复平面上以

0+0i

为圆心、半径为

1

的圆上。所有根的倒数之和必定为

Let

a,

b,

c,

d

be real numbers. Suppose that all the roots of

z^4+az^3+bz^2+cz+d=0

are complex numbers lying on a circle in the complex plane centered at

0+0i

and having radius

1.

The sum of the reciprocals of the roots is necessarily

a

b

c

-a

-b

对于单位圆上的复数

r

，比较

\frac{1}{r}

与

\overline r

For a complex number

r

on the unit circle, compare

\frac{1}{r}

with

\overline r

多项式的系数都是实数，因此所有根的和是实数

Real polynomial coefficients make the sum of the roots real

若

|r|=1

，则

\frac{1}{r}=\overline r

。所以所有根的倒数之和，就是所有根之和的共轭。根据韦达定理，所有根之和为

-a

，这是实数，其共轭仍为

-a

。

因此，正确答案是 D。

If

|r|=1,

then

\frac{1}{r}=\overline r.

Therefore the sum of the reciprocals is the conjugate of the sum of the roots. By Vieta’s formulas, the sum of the roots is

-a,

which is real. Its conjugate is still

-a.

Thus the correct answer is D.

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