Playing with $i$

$\begin{aligned} \left(b+\frac ib \right)^4 & = {\color{#3D99F6}b^4} + 4b^2i - 6 - \frac {4i}{b^2} + {\color{#3D99F6}\frac 1{b^4}} & \small \color{#3D99F6} \text{Note that }b^4 + \frac 1{b^4} = 6 \\ & = 4i \left(b^2 - \frac 1{b^2} \right) & \small \color{#3D99F6} \text{Raise both sides to the power of 4} \\ \left(b+\frac ib \right)^{16} & = 4^4 i^4 \left(b^2 - \frac 1{b^2} \right)^4 \\ & = 4^4 \left[\left(b^2 - \frac 1{b^2} \right)^2\right]^2 \\ & = 4^4 \left[{\color{#3D99F6}b^4} - 2 + {\color{#3D99F6}\frac 1{b^4}} \right]^2 \\ & = 4^4 \cdot 4^2 \\ & = \boxed{4096} \end{aligned}$

Note first that $\left(b + \dfrac{i}{b}\right)^{16} = \left(\left(b + \dfrac{i}{b}\right)^{4}\right)^{4}$ .

Now upon expansion of the binomial, we have that

$\left(b + \dfrac{i}{b}\right)^{4} = b^{4} + 4b^{3} \times \dfrac{i}{b} + 6b^{2} \times \left(\dfrac{i}{b}\right)^{2} + 4b \times \left(\dfrac{i}{b}\right)^{3} + \left(\dfrac{i}{b}\right)^{4} =$

$b^{4} + 4b^{2}i - 6 - \dfrac{4}{b^{2}}i + \dfrac{1}{b^{4}} = 4i\left(b^{2} - \dfrac{1}{b^{2}}\right)$ , since $b^{4} + \dfrac{1}{b^{4}} - 6 = 0$ .

Now $\left(b^{2} - \dfrac{1}{b^{2}}\right)^{2} = b^{4} + \dfrac{1}{b^{4}} - 2 = 6 - 2 = 4$ , so $b^{2} - \dfrac{1}{b^{2}} = \pm 2$ .

Thus $\left(b + \dfrac{i}{b}\right)^{16} = (4i \times (\pm 2))^{4} = (\pm 8i)^{4} = 8^{4} = \boxed{4096}$ .

b^4 + 1/b^4 = 6

b² - 1/b² = 2

b² + i²/b² = 2

Now, ( b + i/b ) ^16 = ( ( b + i/b )² )^8 . = ( b² + i²/b² + 2i )^8 . = ( 2 + 2i)^8 Through binomial expression, we get- . ( 2 + 2i )^8 = 2^8 * 2^4 . = 2^16 =4096

$\begin{aligned}\left(b+\dfrac{i}{b}\right)^{16}&=\left(\sqrt{i}\left(\dfrac{b}{\sqrt{i}}+\dfrac{\sqrt{i}}{b}\right)\right)^{16}\\ &=\left(\dfrac{b}{\sqrt{i}}+\dfrac{\sqrt{i}}{b}\right)^{16}\hspace{5mm}\small\color{#3D99F6}\text{As, }(\sqrt{i})^{16}=1\\ \text{Let,}\\ z&=\left(\dfrac{b}{\sqrt{i}}+\dfrac{\sqrt{i}}{b}\right)\\ z^2-2&=\left(\dfrac{b^2}{i}+\dfrac{i}{b^2}\right)\\ (z^2-2)^2&=-\left(b^4+\dfrac{1}{b^4}\right)+2\\ &=-4\hspace{30mm}\small\color{#3D99F6}\left(b^4+\dfrac{1}{b^4}\right)=6 \text{ Given }\\ \implies z^2&=2\pm2i\\\\ \text{We need}&\text{ to find } z^{16}\\\\ z^{16}&=(z^2)^8=(2\pm2i)^8\\ &=\left(2\sqrt2\hspace{2mm} e^{\pm i\tfrac{\pi}{4}}\right)^8\\ &=\left(2^{12}\hspace{2mm} e^{\pm i2\pi}\right)\\ &=2^{12}\\ &=\color{#EC7300}\boxed{\color{#333333}4096}\end{aligned}$