tylerdesjardins
contestada

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. :)

Respuesta :

[tex]\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}[/tex]

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

Answer:

so, the standard form is:  [tex](\dfrac{13}{3})+i(-3)[/tex]

Step-by-step explanation

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

[tex]a+i b[/tex] where a,b belongs to real numbers.

we are given the expression in complex number as:

[tex]=(3i-\dfrac{2}{3})-(6i-5)[/tex]

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

[tex]=(-\dfrac{2}{3}+5)+i(3-6)\\\\=(\dfrac{13}{3})+i(-3)[/tex]

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

[tex]a=\dfrac{13}{3},b=-3[/tex]