How do I calculate the value of $\tan(1/2)$ for the De Moivre formula I have to solve $(2+i)^3$ using the trigonometric representation. I calculated the modulus but I don't know how to calculate $\varphi$ when it is equal to $\tan(1/2)$.
How do I calculate $\varphi$
Also, is there a fast way to solve these in analogous way if we know that the only thing that changes is modulus power and $\varphi$ product?
$(2+i)^2,\quad (2+i)^3,\quad (2+i)^4,\quad (2+i)^5,\quad (2+i)^6,\quad (2+i)^7$
 A: There is no real advantage in computing powers this way over the algebraic method. There would be an advantage if the argument is a “known angle”.
You can surely write $2+i=\sqrt{5}(\cos\varphi+i\sin\varphi)$, where
$$
\cos\varphi=\frac{2}{\sqrt{5}}\qquad\sin\varphi=\frac{1}{\sqrt{5}}
$$
Then, yes,
$$
(2+i)^3=5\sqrt{5}(\cos3\varphi+i\sin3\varphi)
$$
but now the problem is to compute $\cos3\varphi$ and $\sin3\varphi$, that's no easier than using the binomial theorem from the outset:
$$
(2+i)^3=2^3+3\cdot2^2i+3\cdot2i^2+i^3=8+12i-6-i=2+11i
$$
To wit
$$
\cos3\varphi=\cos^3\varphi-3\cos\varphi\sin^2\varphi
=\frac{8}{5\sqrt{5}}-3\cdot\frac{2}{\sqrt{5}}\frac{1}{5}=\frac{2}{5\sqrt{5}}
$$
and
$$
\sin3\varphi=3\cos^2\varphi\sin\varphi-\sin^3\varphi=
3\cdot\frac{4}{5}\frac{1}{\sqrt{5}}-\frac{1}{5\sqrt{5}}=\frac{11}{5\sqrt{5}}
$$
As you can see, there are exactly the same computations, with added denominators that cancel with the modulus.
A: $2+i  =\sqrt 5 e^{i\theta}$ where $\sin \theta =\frac {1}{\sqrt 5}$ and $ \cos \theta = \frac {2}{\sqrt 5}$
You need $$(2+ i)^3 =5\sqrt 5 (\cos 3\theta + i\sin 3\theta)$$ 
You can figure it out from here. 
A: Outline of steps to solve the problem.


*

*Find $r$ and $\theta$ that makes $re^{i\theta} = 2 + i$

*Notice that $(2 + i)^{3} = (re^{i\theta})^{3} = r^{3}(e^{i\theta})^{3} = r^{3}e^{i3\theta}$

*Find $r^3$. Also Find $\cos(3\theta)+i\sin(3\theta)$, which is $e^{i3\theta}$.

*Find $r^3 \times [\cos(3\theta)+i\sin(3\theta)]$, which is the answer.

