James Ruse 2012 MX2 Trial Q11(d) Geometrical Complex Numbers Question (1 Viewer)

ProdigyInspired · Jul 17, 2016

Now, from i), AOC is found to be $\frac {2\pi}{3}$

What I'm getting confused at is ii). The solution makes sense until they make an expression for $\gamma$
Ok so, $\gamma$ is $\xrightarrow { OC }$ , Arg $\gamma$ is $\theta + \frac {2\pi}{3}$ - which they never use again for some reason.
Then they use pythagoras to say that $|\beta|$ is $\sqrt {2} OA$
how'd they get the expression for $\gamma$ ?

integral95 · Jul 17, 2016

Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question

ProdigyInspired said:
Now, from i), AOC is found to be $\frac {2\pi}{3}$

What I'm getting confused at is ii). The solution makes sense until they make an expression for $\gamma$
Ok so, $\gamma$ is $\xrightarrow { OC }$ , Arg $\gamma$ is $\theta + \frac {2\pi}{3}$ - which they never use again for some reason.
Then they use pythagoras to say that $|\beta|$ is $\sqrt {2} OA$
how'd they get the expression for $\gamma$ ?

Yeah the part that says $\gamma = \beta cis(\frac{2 \pi}{3})$ is incorrect lol, skip that part.

integral95 · Jul 17, 2016

Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question

I would explain by saying that the vector OC (gamma)is OA (alpha) rotated anti-clockwise by 2pi/3 and scaled by root 2

ProdigyInspired · Jul 18, 2016

Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question

So If I do this

$\begin{matrix} \overrightarrow { OA } \end{matrix}rotated\quad by\quad \frac { 2\pi }{ 3 } =\begin{matrix} \overrightarrow { OC } \end{matrix}\\$

How would I go proving the scale?

Would this be right?
$But\quad since\quad the\quad modulus\quad is\quad different\quad i.e.\quad |\gamma |\quad \neq \quad |\alpha |\quad ,\quad \begin{matrix} |\gamma | \end{matrix}is\quad needed\quad to\quad be\quad found\\ \begin{matrix} OC\quad =\quad OB \\ Since\quad AOB\quad is\quad an\quad isosceles\quad right\quad angle\quad triangle, \\ { |\beta | }^{ 2 }\quad =\quad 2\quad \times \quad { |\alpha | }^{ 2 } \\ |\beta |\quad =\quad \sqrt { 2 } \times \quad |\alpha | \\ But\quad |\gamma |\quad =\quad |\beta | \end{matrix}\\ \begin{matrix} \therefore \quad \quad \end{matrix}|\gamma |\quad :\quad |\alpha |\quad =\quad 1\quad :\quad \sqrt { 2 } \\ \therefore \quad \gamma \quad =\quad cis(\frac { 2\pi }{ 3 } )\times \sqrt { 2 } \alpha$

integral95 · Jul 18, 2016

Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question

ProdigyInspired said:
So If I do this

$\begin{matrix} \overrightarrow { OA } \end{matrix}rotated\quad by\quad \frac { 2\pi }{ 3 } =\begin{matrix} \overrightarrow { OC } \end{matrix}\\$

How would I go proving the scale?

Would this be right?
$But\quad since\quad the\quad modulus\quad is\quad different\quad i.e.\quad |\gamma |\quad \neq \quad |\alpha |\quad ,\quad \begin{matrix} |\gamma | \end{matrix}is\quad needed\quad to\quad be\quad found\\ \begin{matrix} OC\quad =\quad OB \\ Since\quad AOB\quad is\quad an\quad isosceles\quad right\quad angle\quad triangle, \\ { |\beta | }^{ 2 }\quad =\quad 2\quad \times \quad { |\alpha | }^{ 2 } \\ |\beta |\quad =\quad \sqrt { 2 } \times \quad |\alpha | \\ But\quad |\gamma |\quad =\quad |\beta | \end{matrix}\\ \begin{matrix} \therefore \quad \quad \end{matrix}|\gamma |\quad :\quad |\alpha |\quad =\quad 1\quad :\quad \sqrt { 2 } \\ \therefore \quad \gamma \quad =\quad cis(\frac { 2\pi }{ 3 } )\times \sqrt { 2 } \alpha$

Yeah that should be fine

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James Ruse 2012 MX2 Trial Q11(d) Geometrical Complex Numbers Question (1 Viewer)

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