2+i/3-i
I am sure you mean
(2+i)/(3-i)
multiply by (3+i)/(3+i)
Since that has a value of 1, we are not changing the value of our original fraction, only its appearance.
(2+i)/(3-i)(3+i)/(3+i)
= (6 + 5i + i^2)/(9-i^2) , recall that i^2 = -1
= (5+5i)/10
= (1+i)/5
yes that is indeed what i meant ...except when you factor
(3+i)(3+i) would it be 9+6i+i^2?
Then what would you do?
Just noticed your reply as I was about to log off ...
Where do you see a multiplication of (x+3)(x+3) ??
Here is what I had:
(2+i)/(3-i)(3+i)/(3+i)
If you follow the order of operation that is
[(2+i)(3+i)] / [(3-i)(3+i}
= as you see above
To simplify the expression (2 + i) / (3 - i), we can start by multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of 3 - i is 3 + i.
So, let's multiply the numerator and denominator by 3 + i:
[(2 + i) * (3 + i)] / [(3 - i) * (3 + i)]
Expanding the numerator and denominator:
(6 + 2i + 3i + i^2) / (9 - 3i + 3i - i^2)
Now, let's simplify the terms:
(6 + 5i + i^2) / (9 + i^2)
Since i^2 is equal to -1, we can substitute this value:
(6 + 5i - 1) / (9 - (-1))
Simplifying further:
(5 + 5i) / 10
To simplify the expression completely, we can divide both the numerator and denominator by 5:
5/10 + (5i)/10
Which simplifies to:
1/2 + (1/2)i
So, the simplified form of (2 + i) / (3 - i) is 1/2 + (1/2)i.