Find the value of a complex number of sin and cos?

joey2
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Homework Statement



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

Homework Equations





The Attempt at a Solution


Our teacher did not really teached me how to do it...
 
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joey2 said:

Homework Statement



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

Homework Equations





The Attempt at a Solution


Our teacher did not really teached me how to do it...

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

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
$$\frac{1+\imath\sqrt{3}}{1+\imath} \\ \text{hint: } \\ 1+\imath\sqrt{3}=2 \left(\frac{1}{2}+\imath\frac{\sqrt{3}}{2}\right)\\1+\imath=\sqrt{2} \left(\frac{\sqrt{2}}{2}+\imath\frac{\sqrt{2}}{2}\right)$$
 
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."
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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