Question

Can someone help me with linear algebra. How do I evaluate this (5-5i)^6

Answer 1

factor out 5, then expand using binomial theorem

Answer 2

notice $$5-5i=5(1-i)=5\sqrt2\left(\frac{\sqrt2}2-\frac{\sqrt2}2i\right)=5\sqrt2\ \left(\cos\left(-\pi/4\right)+i\sin\left(-\pi/4\right)\right)$$

Answer 3

its linear algebra. I have to use de moivres formula

Answer 4

recall de Moivre's theorem which tells us \((\cos t+i\sin t)^n=\cos(nt)+i\sin(nt)\)

Answer 5

yep.

Answer 6

so$$(5-5i)^6=(5\sqrt2\ (\cos(-\pi/4)+i\sin(-\pi/4)))^6=(5\sqrt2)^6(\cos(-\pi/4)+i\sin(-\pi/4))^6$$
and clearly \((5\sqrt2)^6=5^62^3=15625\times8=125000\) whereas $$(\cos(-\pi/4)+i\sin(-\pi/4))^6=\cos(-6\pi/4)+i\sin(-6\pi/4)=\cos(3\pi/2)-i\sin(3\pi/2)$$

Answer 7

now we just recall \(\cos(3\pi/2)=0\) and \(\sin(3\pi/2)=-1\) so:$$(5-5i)^6=125000(0-i)=-125000i$$

Answer 8

ok..my prof gives the answer as (5^6)(2^3)[cos(pi/2) + i sin (pi/2)]

Answer 9

idk how to get the pi.2...i was getting ur answer

Answer 10

The key here is that this applies to vectors on the unit circle. So you have to normalize (5-5i) if you don't automatically recognize that this is just the vector,

\[\frac{ \sqrt{2} }{ 2 }-\frac{ \sqrt{2} }{ 2 }i \] multiplied by a scalar multiple, you can use linear algebra, like the dot product. It might be necessary to do this if it's not a nice angle even multiple of pi.

Answer 11

oops good catch @0213 I had two \(-1\) so it's actually \(+125000i\)

Answer 12

I can elaborate or go into more depth if that sort of lost you. But realize:

cos(x)+isin(x) is just a vector on the unit circle, and you can think of cos(x)*1 as cos(x) being the length and 1 being the unit vector in the x-direction. Then you can think of sin(x) as being the length and i being the unit vector in the y-direction.

Answer 13

See I looked up some videos on this and I get what the videos say and do....however, when I come to do the homework my prof gives me...Idk how to do it or I get the wrong answer. I didn't get the way he was teaching it in class, thats why i looked up the videos and i didn't get why he did some things.....could you explain some things to me and tell me when i should do them

Answer 14

@Kainui

Answer 15

Sure, so what exactly are some of the things that you're unsure of when to do them/ what to do? I'm not really sure where to begin, help me help you. =)

Answer 16

the goal is to write it in the form \(z=r(\cos\theta+i\sin\theta)\) so that we can use de Moivre's theorem i.e. \(z^n=r^n(\cos n\theta+i\sin n\theta)\)

Answer 17

Ok. We just did two lessons so it won't be so long. So the norm? Is that just like the magnitude of the complex number?

Answer 18

@Kainui

Answer 19

the norm is the same thing as the magnitude of course :-)

Answer 20

My prof also said that the argument has to be between 0 and 2pi

Answer 21

but in the videos, they had it over ....what is the difference

Answer 22

your goal is to write it in trigonometric form: