Question

Write (3i - 2/3) - (6i-5) as a complex number in standard form. I have no idea what im doing to be honest -_-. Give me pointers or hints or anything to help me get this done please thanks. :)

Answer 1

`\bf \left( 3i-\cfrac{2}{3} \right)-(6i-5)\implies 3i-\cfrac{2}{3}-6i+5 \\\\\\ -6i+5-\cfrac{2}{3}\implies \begin{array}{llll} \cfrac{13}{3}&-&6i\\ \uparrow &&\uparrow \\ a&-&bi \end{array}`

standard form for a complex expression, I assume it means the a + bi form, nothing else

Answer 2

so, the standard form is:
`((13)/(3))+i(-3)`

Explanation

The standard form of a complex number is given in the form of:

`a+i b` where a,b belongs to real numbers.

we are given the expression in complex number as:

`=(3i-(2)/(3))-(6i-5)`

on combining the like terms i.e. on combining the real and imaginary part of the complex number we get,

`=(-(2)/(3)+5)+i(3-6)\\\\=((13)/(3))+i(-3)`

Hence, on comparing our number with the standard form of the complex number we get

`a=(13)/(3),b=-3`