(D) Given that $f(z) = z^4 + z^3 + 2z^2 + 4z - 8$ has a root $2i$. Since the coefficients are real,the complex conjugate $-2i$ must also be a root.The

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(D) Given that $f(z) = z^4 + z^3 + 2z^2 + 4z - 8$ has a root $2i$. Since the coefficients are real,the complex conjugate $-2i$ must also be a root.Therefore,$(z - 2i)(z + 2i) = (z^2 + 4)$ is a factor of $f(z)$.Dividing $f(z)$ by $(z^2 + 4)$,we get:$f(z) = (z^2 + 4)(z^2 + z - 2)$.Further factoring the quadratic term:$z^2 + z - 2 = (z + 2)(z - 1)$.Thus,the roots of $f(z) = 0$ are $2i, -2i, -2, 1$.Comparing these with the given options,$2$ is not a root of $f(z) = 0$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Mix Examples-Complex numbers

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If $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

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A
$-2i$
B
$1$
C
$-2$
D
$2$

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If $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

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