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

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Homework Help Overview

The discussion revolves around finding the value of a complex number expressed as the division of two complex numbers, specifically \(\frac{1+i\sqrt{3}}{1+i}\). The context includes references to trigonometric functions such as sine and cosine.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods for dividing complex numbers, including using the geometric form of complex numbers and the axiomatic definition of division. There are attempts to clarify the expression and its simplification.

Discussion Status

Some participants have provided hints and suggestions on how to approach the problem, while others express concern that the complexity of the responses may overwhelm the original poster. The discussion remains open with various interpretations being explored.

Contextual Notes

There is an indication that the original poster may lack foundational knowledge on the topic, as noted by their comment about not being taught how to approach the problem. This suggests a potential gap in understanding that is being navigated in the discussion.

joey2
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Homework Statement



fnd 1+i[itex]\sqrt{3}[/itex]/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[itex]\sqrt{3}[/itex]/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
[tex]1 + i \frac{\sqrt{3}}{1} + i = 1 + i \left(\sqrt{3}+1\right).[/tex]
Is that really what you want?
 
I assume you mean you want to find [itex]\frac{1+i\sqrt{3}}{1+i}[/itex]. When it comes to division by complex numbers, we use the strict axiomatic definition that we seldom bother with in real number division: [itex]\frac{a}{b} = a\cdot b^{-1}[/itex] where [itex]b^{-1}[/itex] is the unique number such that [itex]b\cdot b^{-1} = b^{-1}\cdot b = 1[/itex].
So your first job is to find a simpler standard expression for [itex](1 + i)^{-1}[/itex], 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."
 

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