Question

Calculați:Z^2,|Z|,-Z ptr :Z1=1+I,Z2=-i,Z3=-2-i,Z4=3,Z5=-3i-i^2.Dau coroana vă rog repede​

Answer 1

z1=1+i

$z^{2} =(1+i)^{2} =1+2*1*i+i^{2} =1+2i-1=2i\\|z|=|1+i|=\sqrt{1^{2}+1^{2} } =\sqrt{1+1} =\sqrt{2} \\-z=1-i$

(presupun ca acel -z e z conjugat)

z2=-i

$z^{2} =(-i)^{2} =i^{2} =-1\\|z|=|-i|=\sqrt{0 +(-1)^{2}} =\sqrt{1 } =1 \\-z=i$

z3=-2-i

$z^{2} =(-2-i)^{2} =(-2)^{2} -2*(-2)*(-i)+(-i)^{2} =4-4i+i^{2} =4-4i-1=3-4i\\|z|=|-2-i|=\sqrt{(-2)^{2} +(-1)^{2} } =\sqrt{4+1 }= \sqrt{5} \\-z=-2+i$

z4=3

$z^{2} =3^{2} =9\\|z|=|3|=3\\-z=3$

z5=-3i-i^2

$z^{2} =(-3i-i^2)^{2} =(-3)^{2} -2*(-3)*(-i^{2} )+(-i^{2})^{2} =9-6i^{2}+1^{2} =9+6+1=16\\|z|=|-3i-i^2|=|i(-3-i)|=\sqrt{0+(-3-i)^{2} } =\sqrt{9+6i-1} =\sqrt{8+6i} \\-z=3i+i^{2}$

sper ca te-am ajutat