1. danielbarriosr1

$\frac{ 3-i }{ 2+5i }$

2. anonymous

god help me they call everything "simplify" what they mean is "write in standard from as $$a+bi$$" tell your math teacher !

3. danielbarriosr1

Now everything makes sense

4. anonymous

$\frac{ 3-i }{ 2+5i}\times \frac{2-5i}{2-5i}$ is a start

5. danielbarriosr1

the conjugate right?... I don't know if thats how you spell it

6. danielbarriosr1

to eliminate the i in the bottom

7. anonymous

i.e. multiply by the conjugate of the denominator the conjugate of $$a+bi$$ is $$a-bi$$ and this works because $(a+bi)(a-bi)=a^2+b^2$ a real number

8. anonymous

on your case you will have $\frac{(3-i)(2-5i)}{2^2+5^2}$ all the work is now in the numerator

9. anonymous

yeah that is how you spell it too

10. anonymous

you got the top?

11. danielbarriosr1

yeah that would be $\frac{ 4-10i }{ 4-25i^2 }$

12. danielbarriosr1

right? @satellite73

13. danielbarriosr1

My bad, I mean $\frac{ 1-17i }{ 4-25i^2 }$

14. danielbarriosr1

right? @satellite73

15. misty1212

$(a+bi)(a-bi)=a^2+b^2$ so $(2+5i)(2-5i)=2^2+5^2=4+25=29$

16. misty1212

dont mess around with any $$i$$ stuff when you multiply by the conjugate

17. danielbarriosr1

all right so it would be $\frac{ 1 }{ 29 }-\frac{ 17i }{ 29 }$

18. danielbarriosr1

right? @misty1212