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I have done part i) but am having difficulty with part ii)
Thank you
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English Adv | MX1 | MX2 | Phys | Chem | Eco
You can definitely solve it that way, but if the marker is strict, you may not receive full marks for not using the directive "hence".
To do (ii), since you showed that -2i is a root of z^3-8i=0, it follows from the Factor Theorem that (z+2i) is a factor.
Therefore z^3-8i=(z+2i)(z^2+Az-4) , where A can be found from equating coefficients of z^2.
Then find the roots of the quadratic to find the remaining roots of z^3=8i.
Last edited by camelrider; 13 Feb 2018 at 8:27 PM.
I think introducing a new constant is a bit risky, especially because I myself tend to mix up the signs and this method better satisfies the "hence" requirement.
Since:
Therefore:
Hence is a root
Now solve to for the other two roots, you would use the quadratic formula but you can also solve using mod-arg form but you needed to complete the above step to satisfy "hence".
Last edited by darkk_blu; 13 Feb 2018 at 10:05 PM.
as integrand says u can use roots of unity, or factorise z^3=8i to get roots.roots of unity is better thou:
z^3 = 8i
z^3 = 8cis(90)
r^3cis(3x)=8cis(90+360k)
where k is and integer. by adding 360 nothing happens; so all integer value of k allow this condition to be true.
equate r and 8, 3x and 90+360k
r^3=8, r=2
3x=90+360k
x= 30+120k
x= 30, 150,270(-90) , going any further or under for value of k would cause repeats in angles. eg: 390=30
so answer are 2cis30,2cis150,2cis-90
2cis-90 is -2i btw, all the other values also correspond with unique complex numbers
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