Help with Holiday Homework

[zzdfa]

polar form: z^2=18^(1/2)cis(-3PI/4)

thus z=18^(1/4)cis(-3pi/8)

and of course -3pi/8 is the same angle as 5pi/8.

was your answerb ''off'' by 2pi?

edit: fail mistake fixd

[TrueTears]

Let

[darkphoenix]

Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.

[TrueTears]

Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.

Are you using Essentials? They sucked for Complex numbers.

[kyzoo]

Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.

Are you using Essentials? They sucked for Complex numbers.

What's with this "The derivation of this formula is left as an exercise for the reader." So annoying, they never had this in the Methods textbook =(

[TrueTears]

Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.

Are you using Essentials? They sucked for Complex numbers.

What's with this "The derivation of this formula is left as an exercise for the reader." So annoying, they never had this in the Methods textbook =(

lol I see so much of that in the book I'm reading atm.

[darkphoenix]

Nah our school is using the new edition of Maths quest, it sucks.

EDIT: One more question, this on vectors:

If a = 2i - 2j + k and b = -i + j - 4k, find a unit vector perpendicular to a and b.

[Mao]

let c = x i + y j + z k

c . a = 0
2x - 2y + z = 0 [1]

c . b = 0
-x + y - 4z = 0 [2]

(simultaneous equations) 2*[2] + [1]
-7z = 0
z = 0

substituting back into [2] (or [1], will yield the same result)
-x + y = 0
y = x

also, since c is a unit vector,
since z = 0, y = x, substituting yields


Hence or

[darkphoenix]

thanks mao for the quick response.

[kakar0t]

Can someone give me a hand with this question please?

[kamil9876]

a.)

because it is a square. Thus: (1)





b.)

[kakar0t]

Thanks man!