z1=-5+i
z2=1+2i
как решить

z₁ = -5 + i

|z₁| = √((-5)² + 1²) = √26

φ₁ = π - arctg 1/5

z₂ = 1 + 2i

|z₂| = √(1² + 2²) = √5

φ₂ = arctg 2/1 = arctg 2

z=x+iy|z|=✓(x²+y²)z1=-5+ix1=Re z1=-5y1=Im z1=1|z1|=✓((-5)²+1²)=✓26tg(a1)= 1/(-5)= -0,2x1<0y1>0поэтому arg z1=a1=π-arctg(1/5) z1=√26( сos (π-arctg(1/5))+sin( π-arctg(1/5) ))z2=1+2ix2=Re (z2)=1y2=Im (z2)=2|z2|=✓(1²+2²)=✓5arg(z2)=tga=2/1=2x2 >0, y2>0, поэтомуarg z2=a=arctg 2z2=√5( сos (arctg 2)+sin(arctg 2))