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Write (1/3i)-(-6+2/3i) as a complex number in standard form

1 Answer

3 votes

Answer:


6+(i)/(3)

Explanation:


(1)/(3\imath)-(-6+(2)/(3\imath))


(1)/(3\imath)+6-(2)/(3\imath)

taking like terms together


(1)/(3\imath)-(2)/(3\imath)+6

taking LCM


(1-2)/(3\imath)+6


(-1)/(3\imath)+6

taking LCM


(-1+18\imath)/(3\imath)

splitting the term


(-1+18\imath)/(3\imath)

splitting the term


-(1)/(3\imath)+(18\imath)/(3\imath)


-(1*3\imath)/(3\imath * \imath)+6


-(i)/(3\imath^2)+6

we know that


\imath^2=-1

putting this value in above equation


(\imath)/(3)+6

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