Rearranging redshift formula for v

[Pretty Pony]

Hey there
I would just like some help in rearranging the formula:
$$z=\frac{1+\frac{v\cos \theta}{c}}{\sqrt{1-\frac{v^2}{c^2}}}-1<br /> [\tex]<br /> for v.<br /> <br /> Thanks,<br /> PP$$

 

[HallsofIvy]

That is
$$z= \frac{1+ \frac{v cos(\theta)}{c}}{\sqrt{1- \frac{v^2}{c^2}}}- 1$$

You solve an equation by "undoing" what was done to the unknown, step by step. The last thing done there is "-1" so we undo that by adding 1 to both sides:
$$z+ 1= \frac{1+ \frac{v cos(\theta)}{c}}{\sqrt{1- \frac{v^2}{c^2}}}$$

Now, multiply on both sides by the denominator:
$$(z+1)\left(\sqrt{1- \frac{v^2}{c^2}}\right)= 1+ \frac{v cos(\theta)}{c}$$

Get rid of that square root by squaring both sides (which might introduce false solution- you will need to check for that after finding v):
$$(z+1)^2\left(1- \frac{v^2}{c^2}\right)= 1+ 2\frac{v cos(\theta)}{c}+ \frac{v^2 cos^2(\theta)}{c^2}$$

Now that is a quadratic equation in v. Combining like terms,
$$\left(\frac{1}{cos^2(\theta)}+ \frac{(z+1)^2}{c^2}\right)v^2+ \frac{2vcos(\theta)}{c}v+ 1- (z+1)^2= 0$$

which can be solved using the quadratic formula.

 

[Pretty Pony]

Thanks for your help!
:)