Powers of a complex number (1 Viewer)

[mazza_728]

Hey guys .. here is another i cant do! im soo lost

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

Have a go if you can! Please guys


The answer in the book is : 128+128i . I keep getting -64. Can you guys try it...

Thankyou xoxo

 

[KeypadSDM]

Ummm...
(1-i)^3(2+2i)^3
= [(1-i)(2+2i)]^3
= [2 + 2i - 2i - 2i^2]^3
= [2 + 2]^3
= 64

I must have done something wrong ...

Edit: The question was different from what I was attempting.

 

[mazza_728]

Sorry try again: (1-i)^3(2+2i)^4

 

[:: ck ::]

= [ (1-i)(2+2i) ]³ . (2+2i)
= [ (2+2) + i(2 - 2) ]³ . (2+2i)
= 4³ . (2+2i)
= 64 (2+2i)
= 128 + 128 i

 

[mazza_728]

thankyou xoxo

 

[mazza_728]

can u do this the "normal" way ... by that i mean the harder way? because i cant do it!

 

[Supra]

Originally posted by mazza_728
can u do this the "normal" way ... by that i mean the harder way? because i cant do it!

'

wot do u mean normal/harder way??do u mean expanding 1-i sepratly and 2+2i sepratly, then multiplying em??

 

[mazza_728]

you guys are just using simple alegbra which is prolly the smarter method but just for my sanity can someone expand and solve using De Moivre's theorem?

 

[:: ck ::]

hrmmz i don't know what you mean by the normal way

ermmz do u mean converting (1-i) and (2+2i) into polar form and then applying demoivre's theorem?

click below for attachment ~ (soz abt crappy quality.. hopefully still legible)

 

[KeypadSDM]

It's pretty easy when you realise that the arguments are both 45 degrees either way. It makes it VERY easy.

 

[:: ck ::]

yup or in the case or 1-i, 0 - 45^o or -pi/4

wow cool just learned how to write degrees sign in bos

 

[Supra]

Originally posted by :: ryan.cck ::
yup or in the case or 1-i, 0 - 45^o or -pi/4

wow cool just learned how to write degrees sign in bos

how dou write degrees??

 

[:: ck ::]

< sup > o < / sup >

remove any spaces in the tags

 

[Supra]

k thanks, i tried it already and jus deleted the post

 

[mazza_728]

Hey guys wat about this one
If z=3-4i=5(costheta+isintheta) find in x+iy form:
5(sintheta-icostheta)
and
cos2theta+isin2theta

^o²

 

[mazza_728]

sorry it should be ^o4 should it?

Hehe - ive learnt a new thing - thankyou xoxo

 

[ND]

Originally posted by mazza_728
Hey guys wat about this one
If z=3-4i=5(costheta+isintheta) find in x+iy form:
5(sintheta-icostheta)
and
cos2theta+isin2theta

^o²

Have a look at the difference between z and the other two. The first one is very similar to z, only that it's sin@ that's real, and -cos@ is imaginary. Now you have to think: "what can i do to z to get this?" Multiply by i.
The 2nd one is also similar; there difference here is that it's cis2@ instead of 5cis2@. I'm sure you know how to get rid of the 5, and for the 2@ bit you use de moivre's theorem.

 

[mazza_728]

So theres no huge secret that i dont know about? damn i just have to do it the hard way thankyou ND

 

[mazza_728]

AHHH there is too much thinking involved. i understand the second one = (-7-24i / 25) but the first one is really confusing me .. the answer is -4-3i
so you change the i around because the sin is now imaginary but why do we change the sign?

 

[:: ck ::]

umm ...

3 - 4i = 5cis(@)

finding the argument and modulus of 3-4i
|3-4i| = 5
arg(3-4i) = tan^-1(-4/3) = @

------------------------------------------------------------------
a) 5(sin@-icos@)

Taking z = 5(cos@ + isin@)
multiply by i to make sin real and cos imaginary
iz = 5(icos@ -sin@)

now we have sin real and cos imaginary

multiplying by -1
-iz = -5 (-sin(@) + icos(@))
= 5( sin(@) -icos(@) )

which is wot we are after

so if 5(sin@-icos@) = -iz
substituting z = 3-4i
-iz = -i(3-4i)
= -3i + 4i²
.'. .........
(ending with a quote is always good..)

Originally posted by mazza_728
... the answer is -4-3i