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Express `(1-2i)^(-3)`in the standard form `a+i bdot` |
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Answer» Correct Answer - `((-11)/(125)-(2)/(125)i)` `(1-2i)^(-3)=(1)/((1-2i)^(3))=(1)/((1-8i^(3)-6i(1-2i)})` `=(1)/((-11+2i))xx((-11-2i))/((-11-2i))=((-11-2i))/(125)` |
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