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

Question

asked Apr 17, 2020 68.2k views

1 Answer

Mikesl · Answer 1 · 2020-04-23T02:32:16+0000

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

$(\imath)/(3)+6$