$\alpha = 2, \beta=\sqrt3 + i, \gamma=1+i$

Question

Find $r, \theta$ of $\frac{\alpha+\beta}{\gamma}$

For $z=a+bi=re^{i\theta}$, with $ r = \sqrt{a^2+b^2} $ and $\theta= \tan^{-1}\frac{b}{a}$. What is $a,b$ in $\frac{\alpha+\beta}{\gamma}$?

Answer 1

Evaluate,

$$\frac{\alpha+\beta}{\gamma}=\frac{2+\sqrt3+i}{1+i} =(\sqrt3 +1)e^{-i\frac \pi6} $$

where the followings are used,

$$2+\sqrt3+i=\sqrt2(\sqrt3+1)e^{i\frac {\pi}{12}}$$ $$\frac{1}{1+i} = \frac{1}{\sqrt2} e^{-i\frac \pi4}$$

Thus, $r=\sqrt3+1$ and $\theta = -\frac\pi6$.

$$a= \frac{3+\sqrt3}{2},\>\>\>\>\>b= -\frac{1+\sqrt3}{2}$$

——————-

Edit:

To avoid $\pi/12$, evaluate instead

$$\frac{\alpha+\beta}{\gamma} =\frac{2+\sqrt3+i}{1+i}=\frac 12 (2+\sqrt3 +i)(1-i) =\frac 12 [3+\sqrt3 -(\sqrt3+1)i]$$

Then,

$$ r = \frac 12 \sqrt{ (3+\sqrt3)^2+(\sqrt3+1)^2} = \sqrt{4+2\sqrt3}=\sqrt3+1$$

$$\theta =\tan^{-1}-\frac{\sqrt3+1}{3+\sqrt3}=-\tan^{-1}\frac{1}{\sqrt3}=-\frac\pi6$$

Answer 2

After editing the question new answers is

Answer 3

Hint: Multiply the expression $\frac{\alpha + \beta}{\gamma}$ by $\frac{\overline{\gamma}}{\overline{\gamma}}$, where $\overline{\gamma} = 1- i$, and simplify. Note that $\gamma \overline{\gamma}$ is a real number.