User:MathMan64/CentigradeDegree

Fractions

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 $0.\bar{7}$

More examples:

$13.515151...$ or $13.\overline{51}$

$8.6222...$ or $8.6\bar{2}$

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

Changing repeating decimals to fractions

If the entire decimal repeats

Change $0.\bar{7}$ 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 $\varphi =\arctan \left(\frac{b}{a}\right)$. The quadrant of $\varphi$ 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 $0<\varphi <\pi$, 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 $4c^4-4a{c}^2-b^2=0$

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

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