I am having trouble with these problems:
Find all complex numbers $z$ satisfying the equation $$\frac{z+1}{z-1} = i.$$
The value $$\left(\frac{1+\sqrt 3}{2\sqrt 2}+\frac{\sqrt 3-1}{2\sqrt 2}i\right)^{72}$$ is a positive real number. What real number is it?
On 1) , I was thinking about substituting $z$ for $a+bi$ and then solving. Is this correct?
Put one question for each thread please. For number 1, you have $$ i=1+\frac{2}{z-1} \implies z=1+\frac{2}{i-1}=-i. $$
For the first:
We have $$i = 1 + \frac{2}{z-1} \Rightarrow z= -i$$
For the second:
We can write $$\frac{1+\sqrt{3}}{2\sqrt{2}} = \frac{1}{2}\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} = \cos 60^\circ \cos 45^\circ +\sin 60^\circ \sin 45^\circ = \cos (60^\circ -45^\circ ) = \cos 15^\circ$$ Similarly, we have, $\frac{\sqrt{3}-1}{2\sqrt{2}} = \sin 15^\circ$. Thus, using De Moivre's formula, we get, $$(\cos 15^\circ + \sin 15^\circ )^{72} = (\cos 1080^\circ + \sin 1080^\circ) = (\cos 6\pi + \sin 6\pi) = 1$$ Hope it helps.
For 1 just use simple arithmetic: it's a degree one equation.
For 2, compute first the square: \begin{align} \left(\frac{1+\sqrt 3}{2\sqrt 2}+\frac{\sqrt 3-1}{2\sqrt 2}i\right)^{2} &= \left(\frac{1+\sqrt 3}{2\sqrt 2}\right)^2- \left(\frac{\sqrt 3-1}{2\sqrt 2}\right)^2+ 2\frac{1+\sqrt 3}{2\sqrt 2}\frac{\sqrt 3-1}{2\sqrt 2}i \\[6px] &=\frac{1+3+2\sqrt{3}}{8}-\frac{3+1-2\sqrt{3}}{8}+ 2\frac{3-1}{8}i \\[6px] &= \frac{\sqrt{3}}{2}+\frac{1}{2}i \\[6px] &=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6} \end{align}
hint for the first
Your equation can be written as
$$\frac{z-1+2}{z-1}=1+\frac{2}{z-1}=i$$
$$\implies z-1=\frac{2}{(i-1)}=-(i+1)$$
$$\implies z=-i$$
