+1712 \(Express\ \frac{2+i}{4+i}\ in\ the\ form\ a+bi ,\\ where\ a\ and\ b\ are\ real\ numbers.\)
Express (2+i ) /(4+i) in the form a+bi
where a and b are real numbers.
How do I do this, do I use formulas?
I thought you had to just multiply I to the numerator and
to the denominator, but you still have a fraction with I in it...
So what do you do?
EDIT:
What I did:
\(\ \ \ \frac{2+i}{4+i} \\=\frac{(2+i)i}{(4+i)i}\ \\=\frac{2i-1}{4i-1} \\I\ just\ made\ it\ worse.... \\please\ help!\)
-\(tommarvoloriddle \)
+36916 Multiply the given by (4 - i) / (4- i)
= 8 -2i +4i - i^2 / 16 -4i + 4i - i^2
= 8 + 2i +1 / 16 + 1
9 + 2i / (17)
9/17 + 2i/17
+9519 \(\dfrac{2+i}{4+i}\\ = \dfrac{(2+i)(4-i)}{(4+i)(4-i)}\\ = \dfrac{9 + 2i}{4^2 + 1^2}\\ = \dfrac{9}{17} + \dfrac{2}{17}i\)
.
+128408 2 + i
_____ multiply numerator and denominator by the conjugate of 4 + i ⇒ 4 - i
4 + i
2 + i (4 - i) 8 + 4i - 2i - i^2 8 + 2i - (-1) 9 + 2i
____ = ______________ = ___________ = _______ =
4 + i (4 - i) 16 - i^2 16 - (-1) 17
9 2
__ + ___ i
17 17
+2862 This is very simple! Don't be afraid of imaginary numbers guys! If you want to simplify in the form a+bi, like \(\frac{2i+1}{i+3}\), multiply it by the conjugate of the denominator where it is also 1.
So for \(\frac{2i+1}{i+3}\), you multiply it by \(\frac{i-3}{i-3}\), which is the same as 1. This will simplify and won't change the fraction as you are basically multiplying it by 1. Remember to ALWAYS multiply by the conjugate.
This is your method for problems like these
