Divide (2+4i) / (−3+2i)

Answer in standard form for complex numbers.

I attempted this problem, but I'm not sure if it's right. If someone could possibly tell me what I did wrong, then I'd appreciate that. Thank you!

Here's my work:

(2+4i) / (-3+2i) x (-3+2i) / (-3+2i)

= (-6+4i-12+8i^{2}) / (9-6i-6i+4i^{2})

= (-6-8i+8i^{2}) / (9-12i+4i^{2})

= (2i+8(-1)) / (-3+3i+4(-1))

= (2i-8) / (-3+3i-4)

= (8-2i) / (-7+3i).

auxiarc May 13, 2020

#1

#3**-2 **

Guest, when answering questions, it's good practice to include work so the asker won't be like "what the wha"

:)

hugomimihu
May 13, 2020

#2**-2 **

Wellllll

Keep in mind that a+b * a-b = a^2 -b^2

So you should actually multiply by

(2+4i) / (-3+2i) x (-3-2i) / (-3-2i)

And try this out!

I haven't done complex numbrs for a few weeks so I'm kinda rusty.

I'm pretty sure this should work.

I haven't tried it out but I'm reasonably sure that it should work

Update me on any progress!

If you don't understand anything feel free to ask.

hugomimihu May 13, 2020

#4**0 **

Yeah ! If you look at my work, then I did that. I'm just not sure if it's right. That's the only problem. :)

auxiarc
May 13, 2020

#5

#7**-2 **

I believe it should remove the complex number from the denominator and make it un-complex

hugomimihu
May 13, 2020

#8**+3 **

(2 + 4i) / (-3 + 2i) multiply num/den by the conjugate of -3 + 2i = -3 - 2i

So we have

(2 + 4i) ( -3 - 2i)

_____________ =

(-3 + 2i) (-3 - 2i)

-6 - 12i - 4i - 8i^2

_______________ =

9 + 6i - 6i - 4i^2

-6 - 16i + 8

________ =

9 - 4(-1)

2 - 16i

_________ =

9 + 4

(2/13) - (16/13) i

CPhill May 13, 2020

#9**-1 **

Oh Melody's not going to be happy with you

"I'd prefer if you gave hints and not full answers"

LOL

hugomimihu
May 13, 2020