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