The Square of A Complex Number

Observe that:

$\displaystyle \alpha ^{2} = \bigg ( \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}}i \bigg )^{2}$

$\displaystyle \alpha ^{2} = \bigg ( \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}}i \bigg ) \bigg ( \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}}i \bigg )$

Using the distibutive property over addition we get:

$\displaystyle \alpha ^{2} = \frac{1}{ \sqrt{2}} \times \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}} \times \frac{1}{ \sqrt{2}}i + \frac{1}{ \sqrt{2}}i \times \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}}i \times \frac{1}{ \sqrt{2}}i$

$\displaystyle \alpha ^{2} = \frac{1}{2}+ \frac{1}{2}i + \frac{1}{2}i +\frac{1}{2}i^{2}$

Substituting $i^{2} = -1$ :

$\displaystyle \alpha ^{2} = \frac{1}{2}+ \frac{1}{2}i + \frac{1}{2}i - \frac{1}{2}$

$\displaystyle \alpha ^{2} = 2 \bigg (\frac{1}{2}i \bigg)$

$\displaystyle \alpha ^{2} = \frac{2}{2}i$

$\displaystyle \alpha ^{2} = i$

$Clearly,\quad \alpha =cos(\pi /4)+isin(\pi /4)\\ \\ \Rightarrow \alpha ={ e }^{ i(\pi /4) }\\ \\ \therefore { \quad \alpha }^{ 2 }={ { (e }^{ i(\pi /4) } })^{ 2 }={ e }^{ i(\pi /2) }=cos(\pi /2)+isin(\pi /2)=\boxed{i}.$

Solving the square: $(a + b)^2 = a^2 + b^2 + 2ab$

$\alpha^2 = \frac{1}{2} + \frac{1}{2}(-1) + 2*\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}}i$ (note that $2*\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}}$ is equal to 1)

$= \frac{1}{2} - \frac{1}{2} + i = \boxed{i}$

This number is a the first eighth root of unity. We can see that $\alpha^2=\left(e^\frac{2\pi}{8}\right)^2=e^\frac{4\pi}{8},$ which is the second eighth root of unity. The second eighth root of unity is $\boxed{i}$ .

By covert it to polar form:

r=1.

Theta =pi/4.

a^2=(1)^1/2 * (cos (pi/4* 2 )+i sin (pi/4 * 2)).

a^2=1*(cos (pi/2)+i sin (pi/2)).

Now from polar form to standard form:

a^2=1*(0+i).

a^2=i.