How do you find the value of a if $2 i$ $2i$ is a root of $z^{3} - 2 z^{2} i + a z - 32 i = 0$ $z^3-2z^2i+az-32i=0$ ?

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How do you find the value of a if $2 i$ $2i$ is a root of $z^{3} - 2 z^{2} i + a z - 32 i = 0$ $z^3-2z^2i+az-32i=0$ ?

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Answer 1 · 2016-08-12T00:25:36

Hence it is a root just replace the value of $z$ $z$ with $2 i$ $2i$ so we get

${(2 i)}^{3} - 2 {(2 i)}^{2} i + a (2 i) - 32 i = 0 \Rightarrow - 8 i + 8 i + 2 i a - 32 i = 0 \Rightarrow 2 i (a - 16) = 0 \Rightarrow a = 16$ $(2i)^3-2(2i)^2i+a(2i)-32i=0=> -8i+8i+2ia-32i=0=> 2i(a-16)=0=>a=16$

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How do you find the value of a if $2 i$ $2i$ is a root of $z^{3} - 2 z^{2} i + a z - 32 i = 0$ $z^3-2z^2i+az-32i=0$ ?

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How do you find the value of a if 2i2i is a root of z3−2z2i+az−32i=0z^3-2z^2i+az-32i=0?

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How do you find the value of a if $2 i$ $2i$ is a root of $z^{3} - 2 z^{2} i + a z - 32 i = 0$ $z^3-2z^2i+az-32i=0$ ?