7+4i over 2-5i

Question

7+4i

2-5i

 

 

 

I was sick the day my teacher explained this, please help.

Answer 1

(7+4i)/(2-5i)

multiply the numerator and the denominator by the conjugate of (2-5i) which is (2+5i)

(7+4i)(2+5i)/(2-5i)(2+5i)

(14+8i+35i+20i²)/(4-25i²) where i=√-1 and i²=-1

(14+43i-20)/(4+25)

(-6+43i)/29

Answer 2

the trick here is to multiply the top and bottom of the fraction by the conjugate of the denominator

 

if the complex number is a + bi, then the conjugate is   a - bi

 

if the number is -a - bi, the conjugate is -a + bi

 

Notice you only change the sign between the two terms. You don't change all the signs.

 

So for your question, the conjugate of the denominator is   2 + 5i

 

For the numerator you will get (7 + 4i)(2 + 5i) and you will use the FOIL method to multiply.

 

For the denominator you will get (2 - 5i)(2 + 5i) = 4 + 10i - 10i - 25i²  = 4 - 25i²

 

We know i² = -1

 

4 - 25i² = 4 - 25(-1) = 4 + 25 = 29

 

 

Answer 3

Lauren,

 

You must multiply both the top and the bottom of the equation by 2+5i. The purpose is to get rid of "i" in the denominator. When you multiply by the same binomial with the opposite sign in the middle, you eliminate the "i" in the denominator. After you have multiplied the bottom by 2+5i you should get 4-10i+10i+25. This is because you use the foil method to multiply. The denominator is then simplified to 29. The numerator will also be multiplied using the foil method but you will not eliminate the "i" like you did in the denominator. But the important part of solving this is to eliminate the "i" in the denominator.

 

~Jennifer S