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

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data

    fnd 1+i[itex]\sqrt{3}[/itex]/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. jcsd
  3. Apr 22, 2014 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    You wrote
    [tex] 1 + i \frac{\sqrt{3}}{1} + i = 1 + i \left(\sqrt{3}+1\right).[/tex]
    Is that really what you want?
     
  4. Apr 22, 2014 #3
    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.
     
  5. Apr 22, 2014 #4

    lurflurf

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    Homework Helper

    ^?

    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
    $$\frac{1+\imath\sqrt{3}}{1+\imath} \\ \text{hint: } \\ 1+\imath\sqrt{3}=2 \left(\frac{1}{2}+\imath\frac{\sqrt{3}}{2}\right)\\1+\imath=\sqrt{2} \left(\frac{\sqrt{2}}{2}+\imath\frac{\sqrt{2}}{2}\right)$$
     
  6. Apr 22, 2014 #5

    Mark44

    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."
     
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