User:MathMan64/CentigradeDegree

Repeating Decimals
A repeating decimal has digits in the decimal part that repeat forever, such as: $$0.777... \ $$

A shortcut way of writing this is $$0.\overline 7$$

More examples:
 * $$13.515151... \ $$ or $$13.\overline{51}$$
 * $$8.6222... \ $$ or $$8.6\overline 2$$

Notice that the line is only over the part of the decimal that repeats.

If the entire decimal repeats
Change $$0.\overline 7$$ to a fraction.

This one has only one digit that repeats. So multiply by ten.
 * $$0.777... \times 10 = 7.777...$$

Then subtract the original number.
 * $$0.777... \times 10 = 7.777...$$
 * $$0.777... \times 1 \ = 0.777...$$

Subtract in two places: $$10 - 1 = 9 \ $$ and $$ \ 7.777... - 0.777... = 7$$

Square root of a complex number
Each complex number has two square roots. Consider $$z=a+bi \ $$ where $$r=\sqrt{a^2+b^2}$$ and $$ \phi = \arctan \left ( \frac b a \right )$$. The quadrant of $$\phi \ $$ is determined by the signs of a and b.

The square roots are $$\pm\sqrt{\frac{r+a} 2 } \ \pm\ i \ \sqrt{\frac{r-a} 2}$$ where the signs match if $$0< \phi<\pi \ $$, but are different, if not.

This can be derived by expressing $$\sqrt {a+bi} = c+di$$

So $$a+bi=(c+di)^2 = c^2-d^2+2cdi \ $$

Equating the real and imaginary parts: $$a=c^2-d^2 \ $$ and $$b=2cd \ $$

Solve the second equation for d: $$d=\frac b {2c}$$

Substitute into the equation for the real part above to get $$a=c^2-\frac {b^2} {4c^2}$$

Which simplifies to $$4c^4 - 4ac^2 - b^2=0 \ $$

Solving this for $$c^2 \ $$ gives $$\frac{a+\sqrt{a^2+b^2}} 2$$

So $$c=\pm\sqrt{\frac{a+r} 2}$$

Solving $$b=2cd \ $$ for $$c \ $$ and doing similar work gives

$$d=\pm\sqrt{\frac{-a+r} 2}$$