How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ I have these problems :
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$
For some reason this is incorrect I'll be glad to understand why, This is what I done :
I used this formula : $(\alpha,\beta)*(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma)$
And also : $(\alpha,\beta)*(\gamma,\delta)^{-1}=(\alpha,\beta)*(\frac{\gamma}{\gamma^2+\delta^2}-\frac{\delta}{\gamma^2+\delta^2})$


*

*$$\frac{1+i}{1-i}=\\(1+i)(1-i)^{-1}=\\(1+i)(\frac{1}{2}+\frac{1}{2}i)=\\(\frac{1}{2}-\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\1+i\\ \\$$

*$$(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}=\\
((1+\sqrt{3i})(1-\sqrt{3i})^{-1})^{10}=\\
((1+\sqrt{3i})(\frac{1}{1+3i}+\frac{\sqrt{3i}}{1+3i}i)^{10}=\\
(\frac{1}{1+3i}-\frac{\sqrt{3i}\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i})^{10}=\\
(\frac{4}{1+3i}+\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})^{10}$$
Now I want to use De-Moivre :
$$tan(args)= \frac{\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})}{\frac{4}{1+3i}}=\frac{(1+3i)(\sqrt{3i}+\sqrt{3i})}{4+12i}=\frac{\sqrt{3i}-3i\sqrt{3i}+\sqrt{3i}+3i\sqrt{3i}}{4+12i}=\frac{\sqrt{3i}+\sqrt{3i}}{4+12i}$$
But I reach to math error, when trying to calculate the args.
Any help will be appreciated.
 A: What you calculated in the first part is not the same quantity as what you asked in the original question:  is the quantity to be computed $$\frac{1-i}{1+i},$$ or is it $$\frac{1+i}{1-i}?$$  Even if it is the latter, you have made a sign error in the fourth line:  it should be $$(1+i)(\tfrac{1}{2} + \tfrac{i}{2}) = \frac{(1+i)^2}{2} = \frac{1+i+i+i^2}{2} = \frac{2i}{2} = i.$$
For your second question, you need to be absolutely sure that what you mean is $\sqrt{3i}$, rather than $i \sqrt{3}$.  They are not the same.  I suspect the actual question should use the latter, not the former.
A: *

*$$\left(1+i\right)\left(\frac{1}{2}+\frac{1}{2}i\right)\ne\left(\frac{1}{2}\underbrace{\color{red}{\bf -}}_{\small\text{wrong}}\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i\right)$$

*$$((1+\sqrt{3}i)(1-\sqrt{3}i)^{-1})^{10}=
(1+\sqrt{3}i)\left(\frac{1}{\color{red}4}+\frac{\sqrt{3}i}{\color{red}4}\right)^{10}$$

Actually both of them are easily solvable by this way:
- $$\begin{align}\frac{1+i}{1-i}&=\frac{\frac1{\sqrt2}+\frac i{\sqrt2}}{\frac1{\sqrt2}-\frac i{\sqrt2}}\\&=\frac{e^{i\pi/4}}{e^{-i\pi/4}}\\&=e^{i\pi/2}\\&=i\end{align}$$
- $$\begin{align}\left(\frac{1+\sqrt3i}{1-\sqrt3i}\right)^{10}&=\left(\frac{\frac12+\frac{\sqrt3}2i}{\frac12-\frac{\sqrt3}2i}\right)^{10}\\&=\left(\frac{e^{i\pi/3}}{e^{-i\pi/3}}\right)^{10}\\&=\left(e^{2i\pi/3}\right)^{10}\\&=e^{20i\pi/3}=e^{i(6\pi+2\pi/3)}\\&=e^{2i\pi/3}\\&=-\frac12+\frac{\sqrt3}2i\end{align}$$
