Solve the equation ?

Question

Solve the equation ?

Solve the equation ?

${(z - 1)}^{3} = 2 - 11 i$ $(z-1)^3=2-11i$

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Answer 1 · 2018-01-17T12:47:35

$x^{3} - i y^{3} + 3 i x^{2} y - 3 x y^{2} - 3 x^{2} + 3 y^{2} - 6 i x y + 3 x + 3 i y$ $x^3-iy^3+3ix^2y-3xy^2-3x^2+3y^2-6ixy+3x+3iy$ $= 3 - 11 i$ $=3-11i$

Explanation:

given ${(z - 1)}^{3} = 2 - 11 i \to (1)$ $(z-1)^3=2-11i->(1)$ expanding it we get
$z^{3} - 1 - 3 z^{2} + 3 z = 2 - 11 i \to (2)$ $z^3-1-3z^2+3z=2-11i->(2)$
let us assume $z = x + i y \to (3)$ $z=x+iy->(3)$ put $(3)$ $(3)$ in $(2)$ $(2)$
${(x + i y)}^{3} - 3 {(x + i y)}^{2} + 3 (x + i y) = 3 - 11 i$ $(x+iy)^3-3(x+iy)^2+3(x+iy)=3-11i$
$x^{3} - y^{3} i + 3 i x^{2} y - 3 x y^{2} - 3 (x^{2} - y^{2} + 2 i x y) + 3 (x + i y)$ $x^3-y^3i+3ix^2y-3xy^2-3(x^2-y^2+2ixy)+3(x+iy)$ $= 3 - 11 i$ $=3-11i$
$x^{3} - i y^{3} + 3 i x^{2} y - 3 x y^{2} - 3 x^{2} + 3 y^{2} - 6 i x y + 3 x + 3 i y$ $x^3-iy^3+3ix^2y-3xy^2-3x^2+3y^2-6ixy+3x+3iy$ $= 3 - 11 i$ $=3-11i$

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