## Sinz 3 years ago Simplify the following quotient of complex numbers into the form a + bi. -9+8i/1+2i

1. Walleye

You going to mutiply the numorator and the denominatior by the conjugate. Does that help?

2. Sinz

so multiply -9 and 1 right

3. jimmyrep

multiply top and bottom of the quotient by the conjugate of 1 + 2i which is 1 - 2i

4. Sinz

huh I am confused

5. Walleye

Your goal here is to "kill" the bottom complex number. you can do this by \[(9+8i/1+2i) * (1 - 2i / 1 - 2i)\] remember that 1-2i / 1-2i = 1 and you can multiply anything by 1

6. Walleye

without changing the number

7. Sinz

okay

8. jimmyrep

(-9 + 8i)(1-2i) ------------ (1+2i)(1-2i) the denominator simplifies to 1 + 4 = 5

9. Sinz

so I take and combine like terms right?

10. Walleye

yep. you'll find that if you multiply a complex number by its conjugate then you kill all of the i terms and just be left with real numbers

11. Sinz

after I get the )-9+8i)(1-2i)

12. Sinz

so it will be -9*1 and 8i*-2i?

13. jimmyrep

yes - u now simplify the top expression then divide by 5

14. jimmyrep

no remember (a+b)(a+b) = a^2 + 2ab + b^2

15. Sinz

so -9 - 16i/5 is what it should look like right

16. Walleye

nope. your top term is off. Try using FOIL on the top again and see if you get somthing else.

17. Sinz

okay lemme see (-9+8i)(1-2i) combine like terms -9*1=-9 8i*2i=16i total is -9+16i/5 is that right

18. jimmyrep

(-9 + 8i)(1-2i) = -9 + 18i + 8i + 16 = 7 + 26i the second bracket is first multiplied by -9 then its multuplied by + 81

19. jimmyrep

by +8i

20. Sinz

oh okay wasn't seeing how it was done that is why I was confuzzled

21. Walleye

I dont know if this will help but treat i just like you would x when your distributing! I find that thinking about it that way very helpfful

22. jimmyrep

final result is 7/5 + (26/5) i

23. Sinz

okay so in algebra they try to change things up so it confuses you?

24. Sinz

Thank you Jimmy you was very helpful as was you Walleye

25. Walleye

yw!

26. jimmyrep

thats ok