Arithmetic

Problem 3.5.6

a. Given that $13=2^2+3^2$ and $74=5^2+7^2$, express $13 \times 74=962$ as a sum of two squares.

Solution

Let $z=2+3i,w=5+7i$.We have

$$ \begin{align*} 13 \times 74 = |z|^2|w|^2=|zw|^2 = |-11 + 29i| = 11^2+ 29^2 \end{align*} $$

Problem 3.5.8

Show that

$$ \begin{align*} {n \choose 1} -{n \choose 3} + {n \choose 5} - {n \choose 7} + \cdots = 2^{n/2}cos\frac{n\pi}{4} \end{align*} $$

and

$$ \begin{align*} {n \choose 0} -{n \choose 2} + {n \choose 4} - {n \choose 6} + \cdots = 2^{n/2}sin\frac{n\pi}{4} \end{align*} $$

Solution

We have

$$ \begin{align*} (1+i)^n &= {n \choose 0}+ i{n \choose 1} - {n \choose 2} - i{n \choose 3} + \cdots \\
(1-i)^n &= {n \choose 0}- i{n \choose 1} -{n \choose 2} + i{n \choose 3} + \cdots \\
\end{align*} $$

Therefore,

$$ \begin{align*} {n \choose 0} -{n \choose 2} + {n \choose 4} - {n \choose 6} + \cdots &= \frac{(1+i)^n + (1-i)^n}{2} \\ &= 2^{n/2}\frac{e^{in\pi/4} + e^{-in\pi/4}}{2} \\
&= 2^{n/2}cos\frac{n\pi}{4}\\
{n \choose 0} -{n \choose 2} + {n \choose 4} - {n \choose 6} + \cdots &= \frac{(1+i)^n - (1-i)^n}{2} \\&= 2^{n/2}\frac{e^{in\pi/4} - e^{-in\pi/4}}{2} \\
&= 2^{n/2}sin\frac{n\pi}{4} \end{align*} $$

Problem 3.5.9

By considering possible magnitudes and arguments,

a. find all values of $\sqrt[3]{-i}$;

Solution

a. We have $-i= e^{2k\pi - \pi/2}$.Therefore,

$$ \begin{align*} \sqrt[3]{-i}= e^{2k\pi/3 - \pi/6} \end{align*} $$ where $k=0,1,2$.

Problem 3.5.14

Show that if $e^{i\theta}$ satisfies the equation $z^n + a_{n-1}z^{n-1}+\dots+a_1 z+a_0=0$, where the $a_i$ are real, then $a_{n-1}sin\theta + a_{n-2}sin2\theta + \dots + a_1sin(n-1)\theta + a_0sin(n\theta)= 0 $.

Solution

If $e^{i\theta}$ is a solution of the equation, then $e^{-i\theta}$ is also a solution of the equation as the coefficients are real.We have

$$ \begin{align*} a_n + a_{n-1}e^{i\theta} + \dots + a_0e^{in\theta} = 0 \end{align*} $$

Equating the imaginary part of the above equation to zero, we get

$$ \begin{align*} a_{n-1}sin\theta + a_{n-2}sin2\theta + \dots + a_1sin(n-1)\theta + a_0sin(n\theta)= 0 \end{align*} $$