complex numbers help! (1 Viewer)

viraj30 · Nov 24, 2011

the question is- (1) factorise z^5 -1 into real and quadratic factors?? it says (z-1)(1+z+...z^n-1) but i wanna know the steps that ended up with this expression!

pokka · Nov 24, 2011

is (z-1)(1+z+...z^n-1) meant to be a clue? not the answer?

viraj30 · Nov 24, 2011

ok ignore that! it's just factorise z^5 into real and quad factors

tohriffic · Nov 24, 2011

<img src="http://latex.codecogs.com/gif.latex?z^{5} = 1 = 1cis(0)" title="z^{5} = 1 = 1cis(0)" />
<img src="http://latex.codecogs.com/gif.latex?= 1cis(0+2k\pi )" title="= 1cis(0+2k\pi )" /> [because adding 2<img src="http://latex.codecogs.com/gif.latex?\pi" title="\pi" /> is just a full revolution i.e. "bringing you to the same spot" where k is an integer]
<img src="http://latex.codecogs.com/gif.latex?= cis(2k\pi )" title="= cis(2k\pi )" />

Using De Moivre's Thereom
<img src="http://latex.codecogs.com/gif.latex?z = cis(2k\pi )^{\frac{1}{5}}" title="z = cis(2k\pi )^{\frac{1}{5}}" />
<img src="http://latex.codecogs.com/gif.latex?= cis(\frac{2k\pi}{5} )" title="= cis(\frac{2k\pi}{5} )" /> where k = 1, 2, 3, 4 [that's where the n-1 comes in]

Therefore, the five fifth roots are:
<img src="http://latex.codecogs.com/gif.latex?1, cis(\frac{2\pi}{5}), cis(\frac{4\pi}{5}), cis(\frac{6\pi}{5}), cis(\frac{8\pi}{5})" title="1, cis(\frac{2\pi}{5}), cis(\frac{4\pi}{5}), cis(\frac{6\pi}{5}), cis(\frac{8\pi}{5})" /> or in principle argument form: <img src="http://latex.codecogs.com/gif.latex?1, cis(\frac{2\pi}{5}), cis(\frac{4\pi}{5}), cis(\frac{-4\pi}{5}), cis(\frac{-2\pi}{5})" title="1, cis(\frac{2\pi}{5}), cis(\frac{4\pi}{5}), cis(\frac{-4\pi}{5}), cis(\frac{-2\pi}{5})" />

<img src="http://latex.codecogs.com/gif.latex?z^{5} - 1 = (z-1)(z^{4}+z^{3}+z^{2}+ z + 1)" title="z^{5} - 1 = (z-1)(z^{4}+z^{3}+z^{2}+ z + 1)" /> where they represent the roots
<img src="http://latex.codecogs.com/gif.latex?=(z-1)(z-cis(\frac{2\pi}{5}))(z-cis(\frac{4\pi}{5}))(z-cis(\frac{-2\pi}{5}))(z-cis(\frac{-4\pi}{5}))" title="=(z-1)(z-cis(\frac{2\pi}{5}))(z-cis(\frac{4\pi}{5}))(z-cis(\frac{-2\pi}{5}))(z-cis(\frac{-4\pi}{5}))" />
<img src="http://latex.codecogs.com/gif.latex?= (z-1)(z^{2} - 2cos\frac{2\pi}{5}z + 1) (z^{2}-2cos\frac{4\pi}{5}+1)" title="= (z-1)(z^{2} - 2cos\frac{2\pi}{5}z + 1) (z^{2}-2cos\frac{4\pi}{5}+1)" /> [sorry, you're going to have to expand the step before, I didn't type out all my working but I'm sure you know what to do)

Hope this is clear to you.

Don't be afraid to ask questions on my working or anything like that.

bleakarcher · Nov 24, 2011

^ can u show me how u expanded?

tohriffic · Nov 24, 2011

<img src="http://latex.codecogs.com/gif.latex?(z-cis\frac{2\pi}{5})(z-cis\frac{-2\pi}{5})" title="(z-cis\frac{2\pi}{5})(z-cis\frac{-2\pi}{5})" />
<img src="http://latex.codecogs.com/gif.latex?=(z-cos\frac{2\pi}{5}-isin\frac{2\pi}{5})(z-cos\frac{-2\pi}{5}-isin\frac{-2\pi}{5} )" title="=(z-cos\frac{2\pi}{5}-isin\frac{2\pi}{5})(z-cos\frac{-2\pi}{5}-isin\frac{-2\pi}{5} )" />
<img src="http://latex.codecogs.com/gif.latex?=(z-cos\frac{2\pi}{5}-isin\frac{2\pi}{5})(z-cos\frac{2\pi}{5}+isin\frac{2\pi}{5} )" title="=(z-cos\frac{2\pi}{5}-isin\frac{2\pi}{5})(z-cos\frac{2\pi}{5}+isin\frac{2\pi}{5} )" />
<img src="http://latex.codecogs.com/gif.latex?=z^{2}-cos\frac{2\pi}{5}z+zisin\frac{2\pi}{5}-cos\frac{2\pi}{5}z+cos^{2}\frac{2\pi}{5}-isin\frac{2\pi}{5}cos\frac{2\pi}{5}-isin\frac{2\pi}{5}z+isin\frac{2\pi}{5}cos\frac{2\pi}{5}+sin^{2}\frac{2\pi}{5}" title="=z^{2}-cos\frac{2\pi}{5}z+zisin\frac{2\pi}{5}-cos\frac{2\pi}{5}z+cos^{2}\frac{2\pi}{5}-isin\frac{2\pi}{5}cos\frac{2\pi}{5}-isin\frac{2\pi}{5}z+isin\frac{2\pi}{5}cos\frac{2\pi}{5}+sin^{2}\frac{2\pi}{5}" />
<img src="http://latex.codecogs.com/gif.latex?=z^{2}-2cos\frac{2\pi}{5}z+1" title="=z^{2}-2cos\frac{2\pi}{5}z+1" />

And similarly with the other one. Though there is probably a faster way but this is how I did it.

pokka · Nov 24, 2011

$z^5-1=0\\\\z^5=1\\\\$Let $z={\cos\theta} + i{\sin\theta}\\\\$Using De Moivre's Theorem: $z^5=({\cos\theta}+ i{\sin\theta})^5={\cos5\theta}+ i{\sin5\theta}\\\\$and we know that $z^5=1\:$and$\ \cos\2k\pi + i\sin\2k\pi$ where k is an integer, so$\\\\{\cos5\theta}+ i{\sin5\theta}=\cos\2k\pi + i\sin\2k\pi\\\\$i.e$\ 5\theta=2k\pi\\\\\theta=\frac{2k\pi}{5}\\\\k=0,\theta=0,{z_{1}}={\cos\0} + i{\sin\0}=1\\\\k=1,\theta=\frac{2\pi}{5},{z_{2}}={\cos\frac{2\pi}{5}} + i{\sin\frac{2\pi}{5}}\\\\k=2,\theta=\frac{4\pi}{5},{z_{3}}={\cos\frac{4\pi}{5}} + i{\sin\frac{4\pi}{5}}\\\\k=3,\theta=\frac{6\pi}{5},{z_{4}}={\cos\frac{6\pi}{5}} + i{\sin\frac{6\pi}{5}}={\cos\frac{-4\pi}{5}} + i{\sin\frac{-4\pi}{5}}=\overline{z_{3}}\\\\k=4,\theta=\frac{8\pi}{5},{z_{5}}={\cos\frac{8\pi}{5}} + i{\sin\frac{8\pi}{5}}=\cos\frac{-2\pi}{5}} + i{\sin\frac{-2\pi}{5}}=\overline{z_{2}}$ (to be continued...)

pokka · Nov 24, 2011

pokka said:
$z^5-1=0\\\\z^5=1\\\\$Let $z={\cos\theta} + i{\sin\theta}\\\\$Using De Moivre's Theorem: $z^5=({\cos\theta}+ i{\sin\theta})^5={\cos5\theta}+ i{\sin5\theta}\\\\$and we know that $z^5=1\:$and$\ \cos\2k\pi + i\sin\2k\pi$ where k is an integer, so$\\\\{\cos5\theta}+ i{\sin5\theta}=\cos\2k\pi + i\sin\2k\pi\\\\$i.e$\ 5\theta=2k\pi\\\\\theta=\frac{2k\pi}{5}\\\\k=0,\theta=0,{z_{1}}={\cos\0} + i{\sin\0}=1\\\\k=1,\theta=\frac{2\pi}{5},{z_{2}}={\cos\frac{2\pi}{5}} + i{\sin\frac{2\pi}{5}}\\\\k=2,\theta=\frac{4\pi}{5},{z_{3}}={\cos\frac{4\pi}{5}} + i{\sin\frac{4\pi}{5}}\\\\k=3,\theta=\frac{6\pi}{5},{z_{4}}={\cos\frac{6\pi}{5}} + i{\sin\frac{6\pi}{5}}={\cos\frac{-4\pi}{5}} + i{\sin\frac{-4\pi}{5}}=\overline{z_{3}}\\\\k=4,\theta=\frac{8\pi}{5},{z_{5}}={\cos\frac{8\pi}{5}} + i{\sin\frac{8\pi}{5}}=\cos\frac{-2\pi}{5}} + i{\sin\frac{-2\pi}{5}}=\overline{z_{2}}$ (to be continued...)

$\therefore z^5-1\\\\=(z-{z_{1}})(z-{z_{2}})(z-{z_{3}})(z-{z_{4}})(z-{z_{5}})\\\\=(z-1})(z-{z_{2}})(z-{z_{3}})(z-\overline{z_{3}})(z-\overline{z_{2}})$ since $ {z_{1}}=1, {z_{4}}=\overline{z_{3}}$ and ${z_{5}}=\overline{z_{2}}\\\\$and we know that: $ (z-{z_{2}})(z-\overline{z_{2}})=z^2-2\ Re({z_{2}})z+|{z_{2}|^2$ and$\\\\(z-{z_{3}})(z-\overline{z_{3}})=z^2-2\ Re({z_{3}})z+|{z_{3}}|^2\\\\\therefore z^5-1=(z-1)(z^2-2{\cos\frac{2\pi}{5}}z+1)(z^2-2{\cos\frac{4\pi}{5}}z+1)$
ok that took a LONG TIME putting it into LaTex but yeh i hope it is in depth enough to make sense

Alkanes · Nov 24, 2011

^ Good solution. Rep +1

tohriffic · Nov 24, 2011

pokka said:
$\therefore z^5-1\\\\=(z-{z_{1}})(z-{z_{2}})(z-{z_{3}})(z-{z_{4}})(z-{z_{5}})\\\\=(z-1})(z-{z_{2}})(z-{z_{3}})(z-\overline{z_{3}})(z-\overline{z_{2}})$ since $ {z_{1}}=1, {z_{4}}=\overline{z_{3}}$ and ${z_{5}}=\overline{z_{2}}\\\\$and we know that: $ (z-{z_{2}})(z-\overline{z_{2}})=z^2-2\ Re({z_{2}})z+|{z_{2}|^2$ and$\\\\(z-{z_{3}})(z-\overline{z_{3}})=z^2-2\ Re({z_{3}})z+|{z_{3}}|^2\\\\\therefore z^5-1=(z-1)(z^2-2{\cos\frac{2\pi}{5}}z+1)(z^2-2{\cos\frac{4\pi}{5}}z+1)$
ok that took a LONG TIME putting it into LaTex but yeh i hope it is in depth enough to make sense

That's a much simpler solution! Awesome!

complex numbers help! (1 Viewer)

viraj30

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