help with another complex question (1 Viewer)

5uckerberg · Jan 27, 2022

MAK03 said:
View attachment 34805

The first post will be of help for parts i and ii. Part iii will be added later when I had a little bit of thought about it.

part i) Using the properties of a paralleloogram. Already, they have guided you into the question, okay so what do we know about the roots. Have a look at $\alpha, -2\alpha$ that should tell you they are on the same line so already combining that with the properties of a parallelogram we are halfway done. Now we asssemble the other half.

From what we know the property of a parallelogram goes something like this. Once this is shown now we will complete part i. Here, $-3\alpha=\gamma-\alpha+\left(\beta-\alpha\right)=\gamma+\beta-2\alpha$ .
Next, $\alpha+\beta+\gamma=0$

part ii) Ask yourself a question, what is a useful tool from polynomials and part i that we can use to cut out all the noise and immediately show that $\alpha=2i$ ?

Only use the spoilers if you do not know what to do after answering my question

="Use the fact that $\beta + \gamma = -\alpha$ "

Sum of Roots

Lith_30 · Jan 27, 2022

part iii)

Now we just have to find the roots of the equation, so lets just use the identity from part i and the product of roots to simultaneously solve for $\beta$ and $\gamma$ .

$\\\alpha+\beta+\gamma=0\\2i+\beta+\gamma=0\\\beta+\gamma=-2i\\\beta=-\gamma-2i \longrightarrow \textcircled{1}$

Product of roots
$\\\alpha(-2\alpha)\beta\gamma=-32(1+i)\\(2i)(-4i)\beta\gamma=-32(1+i)\\(8)\beta\gamma=-32(1+i)\\\beta\gamma=-4(1+i)\longrightarrow \textcircled{2}$

Now we just sub $\textcircled{1}$ into $\textcircled{2}$

$\\(-\gamma-2i)\gamma=-4-4i\\\--\gamma^2-2i\gamma+4+4i=0$

lets use the quadratic formula

$\\\gamma=\frac{2i\pm\sqrt{(-2i)^2-4(4+4i)}}{-2}\\\\\gamma=\frac{2i\pm\sqrt{{12-16i}}}{-2}\\\\\gamma=\frac{2i\pm2\sqrt{{3-4i}}}{-2}$
Now we can use the identity that the question gave us

$\\\gamma=\frac{2i\pm2(2+i)}{-2}\\\\\gamma=-2-2i,\ \gamma=2$

So yeah the vertices of the parallelogram are -2-2i, 2, 2i, -4i

5uckerberg · Jan 27, 2022

Welp that was helpful @Lith_30, nicely done.

ExtremelyBoredUser · Jan 27, 2022

Lith_30 said:
part iii)

Now we just have to find the roots of the equation, so lets just use the identity from part i and the product of roots to simultaneously solve for $\beta$ and $\gamma$ .

$\\\alpha+\beta+\gamma=0\\2i+\beta+\gamma=0\\\beta+\gamma=-2i\\\beta=-\gamma-2i \longrightarrow \textcircled{1}$

Product of roots
$\\\alpha(-2\alpha)\beta\gamma=-32(1+i)\\(2i)(-4i)\beta\gamma=-32(1+i)\\(8)\beta\gamma=-32(1+i)\\\beta\gamma=-4(1+i)\longrightarrow \textcircled{2}$

Now we just sub $\textcircled{1}$ into $\textcircled{2}$

$\\(-\gamma-2i)\gamma=-4-4i\\\-\gamma^2-2i\gamma+4+4i=0$

lets use the quadratic formula

$\\\gamma=\frac{2i\pm\sqrt{(-2i)^2-4(4+4i)}}{-2}\\\\\gamma=\frac{2i\pm\sqrt{{12-16i}}}{-2}\\\\\gamma=\frac{2i\pm2\sqrt{{3-4i}}}{-2}$
Now we can use the identity that the question gave us

$\\\gamma=\frac{2i\pm2(2+i)}{-2}\\\\\gamma=-2-2i,\ \gamma=2$

So yeah the vertices of the parallelogram are -2-2i, 2, 2i, -4i

the emojis are vital for the proof*

help with another complex question (1 Viewer)

MAK03

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ExtremelyBoredUser

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