April 2 Discussion

The link takes you to this website where some people are constantly monitoring for the next Fermat prime.

http://primes.utm.edu/top20/page.php?id=8

Lenim Depina

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Here's how $\sqrt [ 3 ]{ 2+11i } +\sqrt [ 3 ]{ 2-11i } =4$:

For any complex number in polar form $r(cos\theta +isin\theta )$, or $r(cis\theta )$ for short, that complex number can be brought to any exponent using DeMoivre's Theorem, which states:

(1)

\begin{align} { \left( rcis\theta \right) }^{ n }={ r }^{ n }cis\left( n\theta \right) \end{align}

And this method can be used for complex numbers in rectangular form, $a+bi$, only when it is converted to polar form. To do this, we use two conversion formulas:

(2)

\begin{align} r=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \end{align}

(3)

\begin{align} \theta ={ Arctan }\frac { b }{ a } \end{align}

If we use these conversion formulas, we can turn ${ 2+11i }$ and ${ 2-11i }$ into two equivalent complex numbers in polar form: $\sqrt { 125 } cis\left( Arctan\frac { 11 }{ 2 } \right)$ and $\sqrt { 125 } cis\left( Arctan\frac { -11 }{ 2 } \right)$, respectively. Note that the radical is left unreduced (because it's easier to raise to a fractional power) and the angles are left in terms of the Arctans (because they are not integer angles).

Now, to evaluate the two complex numbers to the $\frac { 1 }{ 3 }$ power, we use DeMoivre's Theorem. Therefore:

(4)

\begin{align} { \left( \sqrt { 125 } cis\left( Arctan\frac { 11 }{ 2 } \right) \right) }^{ \frac { 1 }{ 3 } }={ \sqrt { 125 } }^{ \frac { 1 }{ 3 } }cis\left( \frac { 1 }{ 3 } \cdot Arctan\frac { 11 }{ 2 } \right) \end{align}

(5)

\begin{align} { \left( \sqrt { 125 } cis\left( Arctan\frac { -11 }{ 2 } \right) \right) }^{ \frac { 1 }{ 3 } }={ \sqrt { 125 } }^{ \frac { 1 }{ 3 } }cis\left( \frac { 1 }{ 3 } \cdot Arctan\frac { -11 }{ 2 } \right) \end{align}

When these are turned back into rectangular form (by distribution), they end up
equaling $2+i$ and $2-i$, which when combined equal 4.

Sorry, but what is cis stand for? within those equation
Rachel Bay-Chaparro

• $(cos\theta +isin\theta )$ = $(cis\theta )$