Question

find (3-3i)^6

Answer 1

1) find (3-3i)^2
2) find ((3-3i)^2)^3

Answer 2

Are you solving this using De Moivre's Theorem and stuff like that?
Here is how I would start it off,
\[\Large\bf\sf (3-3\mathcal i)^6\quad=\quad \left(3\sqrt2\right)^6\left(\frac{\sqrt2}{2}-\frac{\sqrt{2}}{2}\mathcal i\right)^6\]
\[\Large\bf\sf =\quad \left(3\sqrt2\right)^6\left(\cos\left[-\frac{\pi}{4}\right]-\mathcal i\sin\left[-\frac{\pi}{4}\right]\right)^6\]

Answer 3

yea

Answer 4

Then applying De Moivre's Theorem,\[\Large\bf\sf =\quad \left(3\sqrt2\right)^6\left(\cos\left[-\frac{6\pi}{4}\right]-\mathcal i\sin\left[-\frac{6\pi}{4}\right]\right)\]Simplifies,\[\Large\bf\sf =\quad \left(3\sqrt2\right)^6\left(\cos\left[-\frac{3\pi}{2}\right]-\mathcal i\sin\left[-\frac{3\pi}{2}\right]\right)\]From there you can simplfy the sine and cosine.

Answer 5

Any confusion on the factoring that I did as the first step?
That part can be a lil tricky.

Answer 6

um naw it seems good

Answer 7

\[\left( 1-\iota \right)^2=1+\iota^2-2\iota=1-1-2\iota=-2\iota \]
\[\left( 1-\iota \right)^6=\left\{ \left( 1-\iota \right)^2 \right\}^3=\left( -2\iota \right)^3=-8*\iota^2*\iota=-8*-1*\iota=8\iota \]
\[\left( 3-3\iota \right)^6=3^6\left( 1-\iota \right)^6=3^6*8\iota=?\]

Answer 8

Ooo I made a boo boo in a few of my steps... blah.
When I converted to sine and cosine, all of those should be + sine, not minus.

Answer 9

Hmm ok good we seem to be getting the same answer,\[\Large\bf\sf 3^6\cdot 8\mathcal i\]

Answer 10

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 11

i think there is something wrong in your second line in the beginning. i may be wrong. i think it should be\[\left( 3\sqrt{2} \right)^6\left\{ \cos (\frac{- \pi }{ 4 })+\iota \sin \left( \frac{- \pi }{ 4 } \right) \right\}^{6}\]

Answer 12

Yes I mentioned that :p