第11章 11.5

(1)

z1=3+iz_1 = 3+i z2=1+2iz_2 = 1+2i z1+z2=4+3iz_1 + z_2 = 4 + 3i

(2)

z1=3−iz_1 = 3-i z2=2+2iz_2 = 2+2i z1−z2=1−3iz_1 - z_2 = 1-3i

(3)

z1=−1+iz_1 = -1+i 2z=−2+2i2z = -2+2i −3z=3−3i-3z = 3-3i

解説: r31bn1z