A few complex number questions (1 Viewer)

cutemouse · Aug 8, 2009

(1) Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2=3$

(2) Prove that for all complex numbers z and w:

$|z+w|^2+|z-w|^2=2[|z|^2+|w|^2|]$

Give a geometrical interpretation of this equation in the complex plane.

(3) (i) If $z=cos\theta+isin\theta$ show that $z^n+\frac{1}{z^n}=2cosn\theta$ (I can do this)
(ii) For $z=r(cos\theta+isin\theta)$ , find r and the smalles value of $\theta$ which may satisfy the equation $2z^3=9+3i\sqrt{3}$ .

(4) Two fixed points $z_1 \ and \ z_2$ and a variable point z represents the complex numbers $z_1 , \ z_2 , \ and \ z$ respectively. Find the locus if:

$arg \frac{z-z_1}{z-z_2} = \beta$

gurmies · Aug 8, 2009

Question 1 (not sure if i've answered the "iff" part correctly)
$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

Question 2
$|z+w|^{2}+|z-w|^{2}=(z+w)\overline{(z+w)}+(z-w)\overline{(z-w)} \\\\ =(z+w)(\overline{z}+\overline{w})+(z-w)(\overline{z}-\overline{w}) \\\\ = z\overline{z}+z\overline{w}+w\overline{z}+w\overline{w}+z\overline{z}-z\overline{w}-w\overline{z}+w\overline{w} \\\\ = 2(z\overline{z}+w\overline{w}) \\\\ = 2(|z|^{2}+|w|^{2})$

I think this means that the sum of the squares of the diagonals of a parallelogram is twice the sum of the squares of the sides?

Aerath · Aug 8, 2009

gurmies said:
I think this means that the sum of the squares of the diagonals of a parallelogram is twice the sum of the squares of the sides?

Yes.

jet · Aug 8, 2009

$2z^3 = 9 + 3\sqrt{3}i \\ r^3 \left (cos 3\, \theta + isin \, 3\theta \right ) = 3\sqrt{3}\left (cos \, \frac{\pi}{6} + isin\, \frac{\pi}{6} \right ) \\ \therefore r^3 = 3\sqrt{3} \Rightarrow r = \sqrt{3} \\ 3\theta = \frac{\pi}{6} + 2k\pi, \text{ k integral} \\ \theta = \frac{\left (12k + 1 \right ) \pi}{18} \\ \therefore z_k = \sqrt{3} \left (cos \left (\frac{\left(12k + 1 \right )\pi}{18} \right ) + isin \left (\frac{\left (12k + 1 \right )\pi}{18} \right ) \right ), k = 0, 1, 2 \\ \\ z_0 = \sqrt{3}\left (cos \frac{\pi}{18}+ isin\frac{\pi}{18} \right )$

$arg \left (\frac{z - z_1}{z - z_2} \right ) = \beta \\ \therefore arg \left (z - z_1 \right ) - arg \left (z - z_2 \right ) = \beta \\ \text{Let } \alpha = arg \left (z - z_! \right ), \gamma = arg \left (z - z_2 \right ) \\ \therefore \alpha - \gamma = \beta \\ \text{Thus, } \alpha = \beta + \gamma \\ \text{We can use the property of a triangle, whereby the exterior angle is equal to the sum of the two interior opposite angles to draw this locus: }$

Note that z₁ and z₂ are completely arbitrary. If they were the other way around, it would be the minor arc between the two of them.
And that is the last math's problem i'll be doing for three weeks.

Aerath · Aug 8, 2009

gurmies said:
Question 1 (not sure if i've answered the "iff" part correctly)
$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

And that is genius.

A few complex number questions (1 Viewer)

cutemouse

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