# Thread: Need Help Simplifying a Few Problems! <Imaginary Numbers>

1. ## Need Help Simplifying a Few Problems! <Imaginary Numbers>

I tried I got the first one I think but i is equal to the square root of -1 so I'm honestly not too sure if it's fully simplified. As for the others I, don't have much of a clue at all.

7 . 9i^26 - 14i^27 / i^7 = 9i^19 - 14i^20

8. 3 - i / 3 + 2i

9. -i (-3i^3 + 4i^2 - 2i + 5) = 3i^4 - 4i^3 + 2i^2 - 5i

10. 2√(-27) - 4√(-75)

2. ## Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

Hey SamBarkley2000.

If i^2 = -1 you can simplify a great deal.

Hint: For Q8 try multiplying by complex conjugate of denominator in the form of (3+2i)/(3+2i) [since it equals 1 it won't change the answer]

3. ## Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

Originally Posted by SamBarkley2000
I tried I got the first one I think but i is equal to the square root of -1 so I'm honestly not too sure if it's fully simplified. As for the others I, don't have much of a clue at all.
7 . 9i^26 - 14i^27 / i^7 = 9i^19 - 14i^20
8. 3 - i / 3 + 2i
9. -i (-3i^3 + 4i^2 - 2i + 5) = 3i^4 - 4i^3 + 2i^2 - 5i
10. 2√(-27) - 4√(-75)
Learn these so you can use them in your sleep.
\begin{align*} i^0 &= 1 \\ i^1 &=i \\i^2 &= -1\\i^3&=-i \end{align*}

$i^{19}=(i^4)^4(i^3)=(1)(i^3)=-i$.

If you divide 20 by 4 the remainder is 0. Thus i^{20}=i^0=1

$i^{74}=-1$ because 74 divided by 4 has a remainder of 2.
and $i^2=-1$.

Also $\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$
Using the above:
$\frac{3-i}{3+2i}=\frac{(3-i)(3-2i)}{9+4}$.

4. ## Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

7. -23

8. 7/13 - 9i/13

9. 1-i

10. -14isqrt(3)

would those be correct chiro, thanks for the hint by the way it helped me out