1. ## Complex number question

Screen Shot 2018-02-13 at 7.56.46 pm.png

I have done part i) but am having difficulty with part ii)
Thank you

2. ## Re: Complex number question

Originally Posted by Karldahemster
Screen Shot 2018-02-13 at 7.56.46 pm.png

I have done part i) but am having difficulty with part ii)
Thank you
$\noindent \textbf{Hint.} First solve using polar form (mod-arg form''). Write z = r\, \mathrm{cis}(\theta) (so our goal is to solve for r and \theta first. See what the equation z^3 = 8i when we substitute z = r\, \mathrm{cis}(\theta), and then compare modulus and argument of both sides of the equation.$

3. ## Re: Complex number question

Originally Posted by InteGrand
$\noindent \textbf{Hint.} First solve using polar form (mod-arg form''). Write z = r\, \mathrm{cis}(\theta) (so our goal is to solve for r and \theta first. See what the equation z^3 = 8i when we substitute z = r\, \mathrm{cis}(\theta), and then compare modulus and argument of both sides of the equation.$
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.

4. ## Re: Complex number question

Originally Posted by camelrider
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.
Oh yeah, I didn't even notice the "hence" (or look at part (i)).

5. ## Re: Complex number question

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.

\begin{aligned}z^3=8i\\z^3-8i=0\end{aligned}

Since:

$(-2i)^3=8i$

Therefore:

\begin{aligned}z^3-8i=0\\z^3-(-2i)^3\\(z-(-2i))(z-(-2i)z+(-2i)^2)=0\\(z+2i)(z+2i-4)=0\end{aligned}

Hence $-2i$ 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".

6. ## Re: Complex number question

$\noindent Sketch of another way to do it via hence'': use the fact that the general solution for z to z^{n}= a^{n} (where a is a given complex number) is z= a\times \left\{n^{\text{th}}\text{ roots of unity}\right\}. So you basically just need to compute the cube roots of unity and multiply them by -2i.$

7. ## Re: Complex number question

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
2cis-90 is -2i btw, all the other values also correspond with unique complex numbers

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