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Math Help - Problem in Complex Number

  1. #1
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    Problem in Complex Number

    Given that z = 5 12i express √z in the form a + bi where a, b Є R.

    How can I expressed this?
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  2. #2
    Senior Member Peritus's Avatar
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    The easyest way of doing this is convert the complex number to polar form then take the square root and convert back to cartesian.
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  3. #3
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    Can you explain more?
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by geton View Post
    Can you explain more?
    sure,
    <br />
\begin{gathered}<br />
  a + ib = \sqrt {a^2  + b^2 } e^{i\arg (a + ib)}  \hfill \\<br />
   \hfill \\<br />
  e^{i\varphi }  = \cos \varphi  + i\sin \varphi  \hfill \\<br />
   \hfill \\<br />
  \sqrt {xe^{iy} }  = \sqrt x e^{i\frac{y}<br />
{2}}  \hfill \\ <br />
\end{gathered} <br /> <br />

    I also strongly suggest that you read:

    Complex number - Wikipedia, the free encyclopedia
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  5. #5
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    mr fantastic's Avatar
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    Quote Originally Posted by geton View Post
    Given that z = 5 – 12i express √z in the form a + bi where a, b Є R.

    How can I expressed this?
    Here's another way:

    Let \sqrt{z} = a + ib \Rightarrow z = (a + ib)^2 = (a^2 - b^2) + i(2ab)

    Therefore 5 - 12i = (a^2 - b^2) + i(2ab).

    Equate real and imaginary parts:

    5 = a^2 - b^2 .... (1)

    -12 = 2ab \Rightarrow -6 = ab \Rightarrow b = -\frac{6}{a} .... (2)

    Sub (2) into (1):

    5 = a^2 - \frac{36}{a^2} \Rightarrow a^4 - 5a^2 - 36 = 0 \Rightarrow (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3.

    The solution a^2 = -4 is rejected because it gives non-real values of a.

    a = 3 \Rightarrow b = -\frac{6}{3} = -2 \Rightarrow \sqrt{z} = 3 - 2i.

    a = -3 \Rightarrow b = -\frac{6}{-3} = 2 \Rightarrow \sqrt{z} = -3 + 2i.

    Given how the numbers fall out, I'd say this was the method the question writer had in mind .....
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  6. #6
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    Quote Originally Posted by Peritus View Post
    sure,
    <br />
\begin{gathered}<br />
  a + ib = \sqrt {a^2  + b^2 } e^{i\arg (a + ib)}  \hfill \\<br />
   \hfill \\<br />
  e^{i\varphi }  = \cos \varphi  + i\sin \varphi  \hfill \\<br />
   \hfill \\<br />
  \sqrt {xe^{iy} }  = \sqrt x e^{i\frac{y}<br />
{2}}  \hfill \\ <br />
\end{gathered} <br /> <br />

    I also strongly suggest that you read:

    Complex number - Wikipedia, the free encyclopedia
    Thank you so much. I got it
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Here's another way:

    Let \sqrt{z} = a + ib \Rightarrow z = (a + ib)^2 = (a^2 - b^2) + i(2ab)

    Therefore 5 - 12i = (a^2 - b^2) + i(2ab).

    Equate real and imaginary parts:

    5 = a^2 - b^2 .... (1)

    -12 = 2ab \Rightarrow -6 = ab \Rightarrow b = -\frac{6}{a} .... (2)

    Sub (2) into (1):

    5 = a^2 - \frac{36}{a^2} \Rightarrow a^4 - 5a^2 - 36 = 0 \Rightarrow (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3.

    The solution a^2 = -4 is rejected because it gives non-real values of a.

    a = 3 \Rightarrow b = -\frac{6}{3} = -2 \Rightarrow \sqrt{z} = 3 - 2i.

    a = -3 \Rightarrow b = -\frac{6}{-3} = 2 \Rightarrow \sqrt{z} = -3 + 2i.

    Given how the numbers fall out, I'd say this was the method the question writer had in mind .....
    Yes I forgot that. Thank you
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