Solution Given: z1+z2=10+4i,z1−z2=6+2iz_1+z_2=10+4i,\qquad z_1-z_2=6+2iz1+z2=10+4i,z1−z2=6+2i Add the equations: (z1+z2)+(z1−z2)=(10+4i)+(6+2i)2z1=16+6iz1=8+3i\begin{aligned} (z_1+z_2)+(z_1-z_2)&=(10+4i)+(6+2i) \\ 2z_1&=16+6i \\ z_1&=8+3i \end{aligned}(z1+z2)+(z1−z2)2z1z1=(10+4i)+(6+2i)=16+6i=8+3i Now substitute into z1+z2=10+4iz_1+z_2=10+4iz1+z2=10+4i: (8+3i)+z2=10+4iz2=2+i\begin{aligned} (8+3i)+z_2&=10+4i \\ z_2&=2+i \end{aligned}(8+3i)+z2z2=10+4i=2+i So: z1=x+yi=8+3i⇒x=8, y=3z_1=x+yi=8+3i\Rightarrow x=8,\ y=3z1=x+yi=8+3i⇒x=8, y=3 z2=a+bi=2+i⇒a=2, b=1z_2=a+bi=2+i\Rightarrow a=2,\ b=1z2=a+bi=2+i⇒a=2, b=1 x=8, y=3, a=2, b=1\boxed{x=8,\ y=3,\ a=2,\ b=1}x=8, y=3, a=2, b=1