complex number multi HSC 2018 (1 Viewer)

[oredbayz]

hi,

is it possible to solve this question without finding all 6 roots and converting them to polar?

thanks

 

[notme123]

hi,

is it possible to solve this question without finding all 6 roots and converting them to polar?

thanks

View attachment 33709

only notable ones here are
just looking at it the only ones that match are A and

CHEESE METHOD.
sub each answer provided into the calculator as . which ever one spits out i, thats your answer

 

[CM_Tutor]

hi,

is it possible to solve this question without finding all 6 roots and converting them to polar?

thanks

View attachment 33709

Yes. The six 6th roots will form a regular hexagon. Further, (C) and (D) have modulus of 4, so can't be roots of i, while (A) and (B) both have modulus of 1 as required. arg(A) is 3pi / 4, so arg(A^6) = 9pi / 2, equiv to pi / 2. arg(B) is 5pi / 4, so arg(B^6) = 15pi / 2, equiv to 3pi / 2. So A^6 = i and B^6 = -i

 

[Siwel]

only notable ones here are
just looking at it the only ones that match are A and
this is the fastest way to do them.

CHEESE METHOD.
sub each answer provided into the calculator as . which ever one spits out i, thats your answer

cant you just put into mod-arg form all of them then find an arg that matches pi/2?

 

[Drongoski]

C) & D) can be discounted immediately because the don't have modulus (length) 1.