\(Express\ \frac{2+i}{4+i}\ in\ the\ form\ a+bi ,\\ where\ a\ and\ b\ are\ real\ numbers.\)

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:

\(\ \ \ \frac{2+i}{4+i} \\=\frac{(2+i)i}{(4+i)i}\ \\=\frac{2i-1}{4i-1} \\I\ just\ made\ it\ worse.... \\please\ help!\)

-\(tommarvoloriddle \)

tommarvoloriddle Jul 13, 2019

#3

#8**0 **

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

ElectricPavlov Jul 13, 2019

#9**+3 **

\(\dfrac{2+i}{4+i}\\ = \dfrac{(2+i)(4-i)}{(4+i)(4-i)}\\ = \dfrac{9 + 2i}{4^2 + 1^2}\\ = \dfrac{9}{17} + \dfrac{2}{17}i\)

.MaxWong Jul 13, 2019

#10**+3 **

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

CPhill Jul 13, 2019

#11**+3 **

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

CalculatorUser Jul 13, 2019

edited by
CalculatorUser
Jul 13, 2019

edited by CalculatorUser Jul 13, 2019

edited by CalculatorUser Jul 13, 2019

edited by CalculatorUser Jul 13, 2019

edited by CalculatorUser Jul 13, 2019