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+8i2) / (9-6i-6i+4i2)
= (-6-8i+8i2) / (9-12i+4i2)
= (2i+8(-1)) / (-3+3i+4(-1))
= (2i-8) / (-3+3i-4)
= (8-2i) / (-7+3i).
Guest, when answering questions, it's good practice to include work so the asker won't be like "what the wha"
:)
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.
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. :)
I believe it should remove the complex number from the denominator and make it un-complex
(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
Oh Melody's not going to be happy with you
"I'd prefer if you gave hints and not full answers"
LOL