complex number multi HSC 2018 (1 Viewer)

oredbayz · Nov 14, 2021

hi,

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

thanks

Screen Shot 2021-11-14 at 10.17.33 am.png

notme123 · Nov 14, 2021

oredbayz said:
hi,

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

thanks

View attachment 33709

$z^6 = i = e^{i\frac{\pi}{2}}, e^{i\frac{-3\pi}{2}}, e^{i\frac{5\pi}{2}}, e^{i\frac{-7\pi}{2}}, e^{i\frac{9\pi}{2}}, e^{i\frac{-11\pi}{2}}$
$z = e^{i\frac{\pi}{12}}, e^{i\frac{-3\pi}{12}}, e^{i\frac{5\pi}{12}}, e^{i\frac{-7\pi}{12}}, e^{i\frac{9\pi}{12}}, e^{i\frac{-11\pi}{12}}$
$z= e^{i\frac{\pi}{12}}, e^{i\frac{-\pi}{4}}, e^{i\frac{5\pi}{12}}, e^{i\frac{-7\pi}{12}}, e^{i\frac{3\pi}{4}}, e^{i\frac{-11\pi}{12}}$
only notable ones here are $e^{i\frac{-\pi}{4}}, e^{i\frac{3\pi}{4}}$
just looking at it the only ones that match are A and $e^{i\frac{3\pi}{4}}$

CHEESE METHOD.
sub each answer provided into the calculator as $z^6$ . which ever one spits out i, thats your answer

CM_Tutor · Nov 14, 2021

oredbayz said:
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 · Nov 14, 2021

notme123 said:
$z^6 = i = e^{i\frac{\pi}{2}}, e^{i\frac{-3\pi}{2}}, e^{i\frac{5\pi}{2}}, e^{i\frac{-7\pi}{2}}, e^{i\frac{9\pi}{2}}, e^{i\frac{-11\pi}{2}}$
$z = e^{i\frac{\pi}{12}}, e^{i\frac{-3\pi}{12}}, e^{i\frac{5\pi}{12}}, e^{i\frac{-7\pi}{12}}, e^{i\frac{9\pi}{12}}, e^{i\frac{-11\pi}{12}}$
$z= e^{i\frac{\pi}{12}}, e^{i\frac{-\pi}{4}}, e^{i\frac{5\pi}{12}}, e^{i\frac{-7\pi}{12}}, e^{i\frac{3\pi}{4}}, e^{i\frac{-11\pi}{12}}$
only notable ones here are $e^{i\frac{-\pi}{4}}, e^{i\frac{3\pi}{4}}$
just looking at it the only ones that match are A and $e^{i\frac{3\pi}{4}}$
this is the fastest way to do them.

CHEESE METHOD.
sub each answer provided into the calculator as $z^6$ . 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 · Nov 14, 2021

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

complex number multi HSC 2018 (1 Viewer)

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