Help Please

In $\left(\frac{1+i}{\sqrt{2}}\right)^{46}$ , we can apply exponent rule, $(\frac{a}{b})^c=\frac{a^c}{b^c}$ . If we do it here, we will get: $\frac{(1+i)^46}{2^{23}}$ .

Find   $\left(\frac{1+i}{\sqrt{2}}\right)^{46}$ .

Let's convert  1 + i  to polar form so we can use de Moivre's Theorem.

$1+i=\sqrt2(\cos\frac{\pi}4+i\sin\frac{\pi}4)$

Now lets raise both sides of that equation to the power of  46 .

$(1+i)^{46}=[\sqrt2(\cos\frac{\pi}4+i\sin\frac{\pi}4)]^{46}\\~\\ (1+i)^{46}=(\sqrt2\,)^{46}(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}$

And divide both sides by  $(\sqrt2\,)^{46}$

$\frac{(1+i)^{46}}{(\sqrt2\,)^{46}}=(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}\\~\\ \left(\frac{1+i}{\sqrt{2}}\right)^{46}=(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}$

De Moivre's Theorem says....

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$    So....

$(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}=\cos(\frac{46\pi}{4})+i\sin(\frac{46\pi}{4})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{46\pi}{4})+i\sin(\frac{46\pi}{4})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{23\pi}{2})+i\sin(\frac{23\pi}{2})$

Now let's find a reference angle that is coterminal with  $\frac{23\pi}{2}$  by subtracting  $2\pi$  five times.

$\frac{23\pi}{2}-2\pi-2\pi-2\pi-2\pi-2\pi\,=\,\frac{23\pi}{2}-10\pi\,=\,\frac{23\pi}{2}-\frac{20\pi}{2}\,=\,\frac{3\pi}{2}$

So....

$\left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=0+i(-1)\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=-i$

Find

$\large{\left(\dfrac{1+i}{\sqrt{2}}\right)^{46}}$

\left(\frac{1+i}{\sqrt{2}}\right)^{46}.

$\begin{array}{|rcll|} \hline && \left(\dfrac{1+i}{\sqrt{2}}\right)^{46} \\\\ &=& \dfrac{ \left(1+i \right)^{46} }{ \left(\sqrt{2} \right)^{46} } \\\\ &=& \dfrac{ \left( \left(1+i \right)^{2}\right)^{23} }{ 2^{\frac{46}{2}} } \quad & | \quad \left(1+i \right)^{2} = 2i \\\\ &=& \dfrac{ \left( 2i \right)^{23} } { 2^{23} } \\\\ &=& \dfrac{ 2^{23}i^{23} } { 2^{23} } \\\\ &=& i^{23} \\\\ &=& \left(i^{2}\right)^{11}i \quad & | \quad i^2 = -1 \\\\ &=& \left(-1 \right)^{11}i \\\\ &\mathbf{=}& \mathbf{-i} \\ \hline \end{array}$