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Vladimir [108]
3 years ago
5

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

Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:

6+\frac{i}{3}

Step-by-step explanation:

\frac{1}{3\imath}-(-6+\frac{2}{3\imath})

\frac{1}{3\imath}+6-\frac{2}{3\imath}

taking like terms together

\frac{1}{3\imath}-\frac{2}{3\imath}+6

taking LCM

\frac{1-2}{3\imath}+6

\frac{-1}{3\imath}+6

taking LCM

\frac{-1+18\imath}{3\imath}

splitting the term

\frac{-1+18\imath}{3\imath}

splitting the term

-\frac{1}{3\imath}+\frac{18\imath}{3\imath}

-\frac{1\times3\imath}{3\imath \times \imath}+6

-\frac{i}{3\imath^2}+6

we know that

\imath^2=-1

putting this value in above equation

\frac{\imath}{3}+6

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Step-by-step explanation:

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