# Homework Help: Find the value of a complex number of sin and cos?

1. Apr 22, 2014

### joey2

1. The problem statement, all variables and given/known data

fnd 1+i$\sqrt{3}$/1+i knowing sin ∏/12 cos ∏/12

2. Relevant equations

3. The attempt at a solution
Our teacher did not really teached me how to do it...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 22, 2014

### Ray Vickson

You wrote
$$1 + i \frac{\sqrt{3}}{1} + i = 1 + i \left(\sqrt{3}+1\right).$$
Is that really what you want?

3. Apr 22, 2014

### slider142

I assume you mean you want to find $\frac{1+i\sqrt{3}}{1+i}$. When it comes to division by complex numbers, we use the strict axiomatic definition that we seldom bother with in real number division: $\frac{a}{b} = a\cdot b^{-1}$ where $b^{-1}$ is the unique number such that $b\cdot b^{-1} = b^{-1}\cdot b = 1$.
So your first job is to find a simpler standard expression for $(1 + i)^{-1}$, and then multiply this number by the numerator. In other words, which complex number do we have to multiply by (1 + i) to get 1 ?
There is, of course, a standardized way to solve this problem that your teacher may have derived in class as well.

4. Apr 22, 2014

### lurflurf

^?

If we write complex numbers in the (geometric) form
$$z=r \,[ \cos(\theta)+\imath \, \sin(\theta)]$$
it is easy to divide using the rule
$$\frac{z_1}{z_2}=\frac{r_1 \,[ \cos(\theta_1)+\imath \, \sin(\theta_1)]}{r_2 \,[ \cos(\theta_2)+\imath \, \sin(\theta_2)]}=\frac{r_1}{r_2} \,[ \cos(\theta_1-\theta_2)+\imath \, \sin(\theta_1-\theta_2)]$$

use this fact to find
$$\frac{1+\imath\sqrt{3}}{1+\imath} \\ \text{hint: } \\ 1+\imath\sqrt{3}=2 \left(\frac{1}{2}+\imath\frac{\sqrt{3}}{2}\right)\\1+\imath=\sqrt{2} \left(\frac{\sqrt{2}}{2}+\imath\frac{\sqrt{2}}{2}\right)$$

5. Apr 22, 2014

### Staff: Mentor

OK guys, no more help until the OP comes back. Also, some of the replies I see here might possibly go right over the head of the OP. You'll have him/her "drinking from a firehose."