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

Answer:

so, the standard form is:  $(\dfrac{13}{3})+i(-3)$

Step-by-step 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:

$=(3i-\dfrac{2}{3})-(6i-5)$

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

$=(-\dfrac{2}{3}+5)+i(3-6)\\\\=(\dfrac{13}{3})+i(-3)$

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

$a=\dfrac{13}{3},b=-3$