Shuckle
  • Shuckle
Write each quotient as a complex number. -2i / 1+i Is -2/-1 and +2i/-1 the correct answer?
Algebra
katieb
  • katieb
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zepdrix
  • zepdrix
\[\large\rm \frac{-2i}{1+i}\left(\frac{1-i}{1-i}\right)=\frac{-2i-2}{2}\]Hmm, do you understand this step that I applied?
Shuckle
  • Shuckle
Yes, I understand that you have to multiply the numerator and denominator by the complex conjugate of the denominator. But from doing that I got the numbers\[\frac{ -2+2i^{2}}{ 1-1i+1i-1^{2}}\]
Shuckle
  • Shuckle
Wow that thing is hard to use. -_-

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zepdrix
  • zepdrix
Woops your denominator is \(\large\rm 1-1i+1i-i^2\)
Shuckle
  • Shuckle
I don't understand?
Shuckle
  • Shuckle
Oh I see, the one is an i. But I still don't know what the final solution should be?
zepdrix
  • zepdrix
So the middle terms cancel out in the denom, ya? :)\[\large\rm \frac{ -2+2i^{2}}{ 1-1i+1i-i^{2}}=\frac{ -2+2i^{2}}{ 1-i^{2}}\]
zepdrix
  • zepdrix
i^2 is -1, so the negatives give us a +1 in that spot
Shuckle
  • Shuckle
Yes, and then multiplied to the denominator of +1 give you a negative one on the bottom?
zepdrix
  • zepdrix
\[\large\rm =\frac{ -2+2(-1)}{ 1-(-1)}\]
zepdrix
  • zepdrix
No you're not multiplying, you're adding down there.
Shuckle
  • Shuckle
Oh.
zepdrix
  • zepdrix
Oh woops that's a 2i for the first term in the numerator :O
zepdrix
  • zepdrix
\[\large\rm =\frac{ -2i+2(-1)}{ 1-(-1)}\]
Shuckle
  • Shuckle
So after that it should be -2i -2 over 1?
zepdrix
  • zepdrix
No, your denominator simplifies to this:\[\large\rm =\frac{ -2i+2(-1)}{ 1+1}\]
Shuckle
  • Shuckle
because (-1)(-) makes a positive one, right?
zepdrix
  • zepdrix
ya :)
Shuckle
  • Shuckle
So then, it'd be -2+2i over 1+1 which will equal -2 +2i over 2, and then I should divide?
Shuckle
  • Shuckle
So the answer is -1-i?
zepdrix
  • zepdrix
yayyy good job \c:/
Shuckle
  • Shuckle
Thank you so much. :3

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