If the roots of the equation Z^3+i Z^2+2 i=0 | vedclass

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(C) Given the equation $Z^3+i Z^2+2 i=0$. By inspection,$Z=i$ is a root because $i^3+i(i^2)+2i = -i-i+2i = 0$. Dividing the polynomial by $(Z-i)$,we g

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an isosceles triangle

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(C) Given the equation $Z^3+i Z^2+2 i=0$. By inspection,$Z=i$ is a root because $i^3+i(i^2)+2i = -i-i+2i = 0$. Dividing the polynomial by $(Z-i)$,we get $(Z-i)(Z^2+2iZ-2)=0$. Solving $Z^2+2iZ-2=0$ using the quadratic formula: $Z = \frac{-2i \pm \sqrt{(2i)^2 - 4(1)(-2)}}{2} = \frac{-2i \pm \sqrt{-4+8}}{2} = \frac{-2i \pm 2}{2} = -i \pm 1$. The roots are $Z_1 = i$,$Z_2 = 1-i$,and $Z_3 = -1-i$. Representing these as points in the Argand plane: $A(0, 1)$,$B(1, -1)$,and $C(-1, -1)$. Calculating the side lengths: $AB = \sqrt{(1-0)^2 + (-1-1)^2} = \sqrt{1+4} = \sqrt{5}$. $BC = \sqrt{(-1-1)^2 + (-1-(-1))^2} = \sqrt{(-2)^2 + 0} = 2$. $AC = \sqrt{(-1-0)^2 + (-1-1)^2} = \sqrt{1+4} = \sqrt{5}$. Since $AB = AC = \sqrt{5}$,the triangle is an isosceles triangle.

If the roots of the equation $Z^3+i Z^2+2 i=0$ are the vertices of a triangle $ABC$,then that triangle $ABC$ is

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