1. myininaya

so you have 3+2i/3-2i or do you mean (3+2i)/(3-2i)?

2. myininaya

i think you mean the latter $\text{ so we have } \frac{3+2i}{3-2i}$ do you know to put it in a+bi form the first step you must take is to multiply both top and bottom by bottom's conjugate

3. myininaya

do you know what the conjugate of 3-2i is?

4. spandana

multipy numerator and denominator with 3+2i

5. amyna

@myininaya yes i multiplied the top and bottom by 3+2i and got 9+6i+6i+4i^2 over 9+6i-6i-4i^2 and got the final answer as 9+12i / 9 but its the wrong answer

6. myininaya

so how did you get 9+4i^2 is 9?

7. myininaya

recall i^2 is -1

8. myininaya

so you got it right so far to say $\frac{9+6i+6i+4i^2}{9+6i-6i-4i^2}$

9. myininaya

and you know again i^2 is -1 so you have $\frac{9+12i+4(-1)}{9-4(-1)}$ wherever there was an i^2 i just replaced it with -1

10. myininaya

simplify this and you will see you got the answer as it should be written

11. amyna

Thanks so much @myininaya

12. myininaya

I think this means you got it so gj