A couple of questions (1 Viewer)

[Riviet]

I need some help with these:
Evaluate in the form a+bi, by using De Moivre's theorem [2(cos2pi/3-isin2pi/3)]⁴ and it had a hint which i didn't get "bracketed expression=cis-2pi/3"

Express in mod-arg form and cartesian form (3+4i)^-2
I got 1/25(cis-106^o) but answers had something completely different.
Please show lots of working, thx

 

[KFunk]

I need some help with these:
Evaluate in the form a+bi, by using De Moivre's theorem [2(cos2pi/3-isin2pi/3)]⁴ and it had a hint which i didn't get "bracketed expression=cis-2pi/3"

What they meant in the hint is that cos(2π/3) - isin(2π/3) = cos(-2π/3) + isin(-2π/3) which puts it in the form cos@ + isin@ which you can apply de moivre's to since (cis@ + isin@)ⁿ = cis(n@) + isin(n@).

The hint works because cos is an even function: cosx = cos(-x) and because sin is an odd function: sin(-x) = -sinx

Give it a go using the hint and given that [r(cos@ + isin@)]ⁿ = rⁿ[cos(n@) + isin(n@)]

 

[Antwan23q]

ohhh
well just so u know
R[cos@+isin@] can be written in shorthand Rcis@
where c means the cos, is is the isin

so anyways
[2(cos2pi/3-isin2pi/3)]^4 = 2^4cis(2pi/3x4)
=16cis(8pi/3) because cos(8pi/3)=cos(2pi/3)
=16cis(2pi/3)

 

[dawso]

1 - simple demoivres theorem.....if u dont know this then u hav alot of work before monday - 16 cis 8pi/3

2 - first get 3+4i in mod arg form....5cis 53 i think it is from memory, then do the same as above




on second thoughts, you have to be a 2006 hscer?

 

[icycloud]

on second thoughts, you have to be a 2006 hscer?

It appears that he is .

 

[Antwan23q]

he is 2006er

 

[Riviet]

Thank-you-thank-you KFunk!!! I get it! I get it!








<-------------Yes, i'm an 06er of course