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 $0.{\overline {7}}$

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.

Changing repeating decimals to fractions

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

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