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.

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.
