8-i/2+3i
write complex # in standard form.
I know to FOIL, not sure if I did it right,
I got -13/4-9i then I'm not sure what to do next or if I'm done...so lost can anyone help me?
I will assume that you mean
(8 - i)/(2 + 3i) ....... the brackets are crucial here
= (8 - i)/(2 + 3i) * (2-3i)/(2-3i_
= (16 - 24i - 2i + 3i^2)/(4 + 9)
= (13 - 26i)/13
= 1 - 2i
Thanks for the help Reiny!
To write the complex number 8 - i/2 + 3i in standard form, we need to combine the real and imaginary parts.
Let's start by expressing the given expression without any fractions. To achieve this, we can multiply the entire expression by 2 to get rid of the denominator:
8 - i/2 + 3i (multiply by 2)
= 16 - i + 6i
Next, combine like terms:
= (16 + 6i) - i
= 16 + 6i - i
= 16 + 5i
Now the complex number is in the form a + bi, where a represents the real part and b represents the imaginary part.
Therefore, the complex number 8 - i/2 + 3i in standard form is 16 + 5i.