Complex Number Problem

[Grace Wilkerson]

I worked this problem and came up with the answer -1/2 + 2i and this is different than the answer in the book -1/2 - 2i. I think the book is wrong. Can you help?

8i⁴ - 4i³ - 6i
-3i³ + 5i - square root of -16

Here's my work:

8 + 4i - 6i
3i + 5i - 4i

8 - 2i
4i

8 2i
4i - 4i

2i - 1/2 = -1/2 + 2i

 

[pka]

I worked this problem and came up with the answer -1/2 + 2i and this is different than the answer in the book -1/2 - 2i. I think the book is wrong. Can you help?

\(\displaystyle \dfrac{8-2i}{4i}=\dfrac{8-2i}{4i}\left(\dfrac{-4i}{-4i}\right)=\dfrac{-8-32i}{16}=\dfrac{-1-4i}{2}\).

 

[Grace Wilkerson]

Thank you.

\(\displaystyle \dfrac{8-2i}{4i}=\dfrac{8-2i}{4i}\left(\dfrac{-4i}{-4i}\right)=\dfrac{-8-32i}{16}=\dfrac{-1-4i}{2}\).

Thank you for your help. I see where I went wrong.

 

[Grace Wilkerson]

Thank you

\(\displaystyle
\dfrac{8i^4-4i^3-6i}{-3i^3+5i -\sqrt{-16}} = \)

\(\displaystyle
\dfrac{8i^4 - 4i^3 - 6i}{3i + 5i - 4i} =\)

\(\displaystyle \dfrac{8i^3-4i^2-6}{3+5-4} = \)

\(\displaystyle \dfrac{-8i + 4 - 6}{4} =\)

\(\displaystyle \dfrac{-8i - 2}{4} =\)

\(\displaystyle -2i -\dfrac{1}{2} \)

The book looks right to me.

Thank you for your help. I see where I went wrong.