Question

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

Answer 1

`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`