1.2 參考解答

  1. (a) Since z1z_1 is a root of polynomial PP, we have P(z1)=0P(z_1)=0.

And P(z1‾)‾=z1‾n+az−1z1‾n−1+⋯+a1z1‾+a0‾\overline{P(\overline{z_1})}=\overline{\overline{z_1}^n+a_{z-1}\overline{z_1}^{n-1}+\cdots+a_1\overline{z_1}+a_0}

=z1‾n‾+an−1z1‾n−1‾+⋯+a1z1‾‾+a0‾=\overline{\overline{z_1}^n}+\overline{a_{n-1}\overline{z_1}^{n-1}}+\cdots+\overline{a_1\overline{z_1}}+\overline{a_0}

=z1‾‾n+an−1z1‾‾n−1+⋯+a1z1‾‾+a0=\overline{\overline{z_1}}^n+a_{n-1}\overline{\overline{z_1}}^{n-1}+\cdots+a_1\overline{\overline{z_1}}+a_0

=z1n+an−1z1n−1+⋯+a1z1+a0=P(z1)=z_1^n+a_{n-1}z_1^{n-1}+\cdots+a_1z_1+a_0=P(z_1)

Then we have P(z1‾)‾=0\overline{P(\overline{z_1})}=0 and P(z1)‾‾=0\overline{\overline{P(z_1)}}=0.

i.e. P(z1‾)=0P(\overline{z_1})=0.

Thus, z1‾\overline{z_1} is a root of polynomial PP.

  1. (b) False

counterexa奧mple:

Let z1=1+iz_1=1+i, z2=1−iz_2=1-i

z1⋅z2=(1+i)(1−i)=1−(−1)=2 z_1 \cdot z_2=(1+i)(1-i)=1-(-1)=2

Re(z1z2)=2\text{Re}(z_1z_2)=2

Re(z1)=1\text{Re}(z_1)=1

Re(z2)=1\text{Re}(z_2)=1

Re(z1)Re(z2)=1\text{Re}(z_1)\text{Re}(z_2)=1

∴Re(z1z2)=2≠Re(z1)Re(z2)\therefore \text{Re}(z_1z_2)=2 \neq \text{Re}(z_1)\text{Re}(z_2)

  1. Let z1=(x1,y1)z_1=(x_1,y_1), z2=(x2,y2)z_2=(x_2,y_2), z3=(x3,y3)z_3=(x_3,y_3)

z1(z2+z3)=(x1,y1)[(x2,y2)+(x3,y3)]z_1(z_2+z_3)=(x_1,y_1)[(x_2,y_2)+(x_3,y_3)]

=(x1,y1)(x2+x3,y2+y3)=(x_1,y_1)(x_2+x_3,y_2+y_3)

=(x1(x2+x3)−y1(y2+y3),x1(y2+y3)+y1(x2+x3))=(x_1(x_2+x_3)-y_1(y_2+y_3), x_1(y_2+y_3)+y_1(x_2+x_3))

=(x1x2+x1x3−y1y2−y1y3,x1y2+x1y3+y1x2+y1x3)=(x_1x_2+x_1x_3-y_1y_2-y_1y_3, x_1y_2+x_1y_3+y_1x_2+y_1x_3)

=(x1x2−y1y2,x1y2+y1x2)+(x1x3−y1y3,x1y3+y1x3)=(x_1x_2-y_1y_2, x_1y_2+y_1x_2)+(x_1x_3-y_1y_3, x_1y_3+y_1x_3)

=(x1,y1)(x2,y2)+(x1,y1)(x3,y3)=(x_1,y_1)(x_2,y_2)+(x_1,y_1)(x_3,y_3)

=z1z2+z1z3=z_1z_2+z_1z_3