complex evaluation (1 Viewer)

hornsbystudylad · Nov 21, 2022

help with c)ii)?

Lith_30 · Nov 21, 2022

using the diagram from part i or just by solving for the roots of unity $z_3=cis\left(\frac{2\pi}{3}\right)$ and $z_2=cis\left(-\frac{2\pi}{3}\right)$ and $z_1=1$

now the expression $|z_1-1|^2+|z_2-1|^2+|z_3-1|^2$ can be simplified to $|z_2-1|^2+|z_3-1|^2$ cause $z_1=1$

then we just sub in $z_2$ and $z_3$ in

$\\\left|-\frac{1}{2} + \frac{\sqrt{3}i}{2}-1\right|^2+\left|-\frac{1}{2}-\frac{\sqrt{3}i}{2}-1\right|^2\\\\=2\left(\frac{3}{2}^2+\frac{\sqrt{3}}{2}^2\right)\\\\=6$

Drongoski · Nov 21, 2022

$z_1 \text{ not necessarily } = 1$ but answer still = 6.

$z_1 = cis\theta; $ $ z_2 = cis(\theta + \frac{2\pi}{3}); $ $ z_3 = cis(\theta + \frac {4\pi}{3})\\ \\ |z_1 -1|^2 + |z_2 - 1|^2 + |z_3-1|^2 = (z_1-1)\overline{(z_1 - 1)} + (z_2 - 1)\overline{(z_2 - 1)} + (z_3 - 1)\overline{(z_3 - 1)}\\ \\ = (z_1-1)(\overline{z_1}-1) + (z_2 - 1)(\overline{z_2}-1) + (z_3)(\overline {z_3}-1)\\ \\=( z_1\overline{z_1} -2(z_1+ \overline{z_1} )+ 1) + (z_2 \overline{z_2} - 2(z_2 + \overline {z_2}) +1) + (z_3 \overline{z_3} -2(z_3 + \overline {z_3}) +1)\\ \\ = (1-2cos\theta + 1) + (1- 2cos(\theta + \frac {2\pi}{3}) +1) + (1-2 cos(\theta + \frac {4\pi}{3})+1)\\ \\ = 6 - 2(cos\theta + [cos\theta cos\frac {2\pi}{3} - sin\theta sin \frac{2\pi}{3}] +[cos\theta cos \frac {4\pi}{3} - sin \theta sin \frac {4\pi}{3}])\\ \\ = 6 - 2(cos \theta - \frac {1}{2} cos\theta -\frac {\sqrt 3}{2}sin\theta -\frac{1}{2}cos \theta + \frac {\sqrt 3}{2} sin \theta)\\ \\ = 6 - 2(0) = 6$

So the answer is independent of the value of $\theta$

Lith_30 · Nov 21, 2022

Drongoski said:
$z_1 \text{ not necessarily } = 1$ but answer still = 6.

what else could $z_1$ equal to?

chilli 412 · Nov 21, 2022

Lith_30 said:
what else could $z_1$ equal to?

Screen Shot 2022-11-21 at 12.55.16 pm.png

i was playing around with this on desmos and found out z1 can actually be an infinite number of complex numbers with argument less than pi/2 , all the complex numbers have a modulus of 1 as well, here is the desmos link (i spent a long time doing this in business studies) desmos link

Drongoski · Nov 21, 2022

Lith_30 said:
what else could $z_1$ equal to?

$z_1 = cis \theta \cdots (0 \leq \theta \leq \frac {\pi}{2})$

You can argue geometrically that the sum of the squares of the distances remains unchanged (is invariant) when we rotate the figure anticlockwise until theta = 0 (your special case) or any other angle.

Drongoski · Nov 21, 2022

chilli 412 said:
View attachment 37090
i was playing around with this on desmos and found out z1 can actually be an infinite number of complex numbers with argument less than pi/2 , all the complex numbers have a modulus of 1 as well, here is the desmos link (i spent a long time doing this in business studies) desmos link

Nice diagram. But the 3 red lines must be equal in length, being radii of the unit circle.

chilli 412 · Nov 21, 2022

Drongoski said:
Nice diagram. But the 3 red lines must be equal in length.

they should all have a length of 1

hornsbystudylad · Nov 21, 2022

thanks kings

Lith_30 · Nov 21, 2022

chilli 412 said:
View attachment 37090
i was playing around with this on desmos and found out z1 can actually be an infinite number of complex numbers with argument less than pi/2 , all the complex numbers have a modulus of 1 as well, here is the desmos link (i spent a long time doing this in business studies) desmos link

interesting graph, have a look at this graph which probably shows the question better and also helps demonstrate what @Drongoski was saying about the anser being independent of the argument of z1.

chilli 412 · Nov 21, 2022

Lith_30 said:
interesting graph, have a look at this graph which probably shows the question better and also helps demonstrate what @Drongoski was saying about the anser being independent of the argument of z1.

oh woops i just tested my one with a circle of radius 1 and they arent all one ill try to fix it

complex evaluation (1 Viewer)

hornsbystudylad

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