Question is attatched. This is word for word from the question, including the "real real" bit.. which doesnt actually make any sense to me, a typo? How do you resolve z4+1 = 0 into real (imaginary?) quadratic factors?
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Factorizing Complex Equations (1 Viewer)
- Thread starter Mumma
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I think the question was meant to have 1 "real" lol.
Anyway this factorisation is a little tricky.
z4+1=0
(z2+1)2-(sqrt2z)2=0
(z2+sqrt(2z)+1)(z2-sqrt(2z)+1)=0
This is now factorised into "real quadratic factors"
Now using quadratic formula for each quadratic:
z=[-sqrt2+/-sqrt2(i)]/2, [sqrt2+/-sqrt2(i)]/2
You can graph these on the arg diagram for yourself. Working on deducing bit now.
Anyway this factorisation is a little tricky.
z4+1=0
(z2+1)2-(sqrt2z)2=0
(z2+sqrt(2z)+1)(z2-sqrt(2z)+1)=0
This is now factorised into "real quadratic factors"
Now using quadratic formula for each quadratic:
z=[-sqrt2+/-sqrt2(i)]/2, [sqrt2+/-sqrt2(i)]/2
You can graph these on the arg diagram for yourself. Working on deducing bit now.
Yeah I got the deduce bit. Friggin easy trig. It seems to have absolutely nothing to do with z4+1 at all. Just expand RHS and it simplifies down to cos2(Theta)
BTW I got the first part before you replied, but thank you anyway!
How I did the first part was I found the roots of z4+1 (using z = cis[(360k+180)/2],
put them in the form (z-z1)(z-z2)(z-z3)(z-z4),
expanded and simplified down to z4+1 = (z2-zsqrt2+1)(z2+zsqrt2+1)
I like the completing the square method though, thanks for that
BTW I got the first part before you replied, but thank you anyway!
How I did the first part was I found the roots of z4+1 (using z = cis[(360k+180)/2],
put them in the form (z-z1)(z-z2)(z-z3)(z-z4),
expanded and simplified down to z4+1 = (z2-zsqrt2+1)(z2+zsqrt2+1)
I like the completing the square method though, thanks for that
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