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Math Help - complex number

  1. #1
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    complex number

    Given that z=-2+i , express  z+\frac{1}{\sqrt{z}} in the form of a+bi
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  2. #2
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    Write -2 + i as x + i y ; r = (x^2 + y^2)^ (1/2)

    Z = x + i y = r e^ i theta = r ( cos theta + i sin theta) from De Moivre;s
    Theorem


    where sin theta = y / (x^2 + y^2)^ (1/2) and

    cos theta = x / (x^2 + y^2)^ (1/2)

    Just substitute for Z + 1/Z and remember sin(-theta) = - sin theta.
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  3. #3
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    Is there a simpler way of doing this ?

    This is what i tried :

    \sqrt{-2+i}=a+bi

    -2+i=a^2-b^2+2abi

    -2=a^2-b^2 and 2ab=1

    i am not sure whether i am correct up to this step but after solving , i get weird values for a and b ..
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  4. #4
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    Actually the problem is because there are roots of a complex no involved I think and we have to involve the unit circle etc.

    I tried it too . I got 4a^4 + 8a^2 - 1 = 0 . Put a^2 = x and then solve

    4x^2 + 8 x - 1 = 0 to get x = -1 +or- 1/2 sqrt 5

    Choose the +ve value and x = 0.118 or a = 0.3435 . Similarly try for b.
    We have to ignore the value which is -ve else we end up with a being a complex no starting from the assumption it is not!
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  5. #5
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    Talking

    Quote Originally Posted by thereddevils View Post
    Given that z=-2+i , express  z+\frac{1}{\sqrt{z}} in the form of a+bi
    Just do the simplification:

    -2\, +\, i\, +\, \frac{1}{-2\, +\, i}

    -2\, +\, i\, +\, \left(\frac{1}{-2\, +\, i}\right)\left(\frac{-2\, -\, i}{-2\, -\, i}\right)

    -2\, +\, i\, +\, \frac{-2\, -\, i}{(-2)^2\, -\, i^2}

    -2\, +\, i\, +\, \frac{-2\, -\, i}{4\, +\, 1}

    -2\, +\, i\, +\, \frac{-2\, -\, i}{5}

    -2\, -\, \frac{2}{5}\, +\, i\, -\, \frac{1}{5}i

    -\frac{12}{5}\, +\, \frac{4}{5}i

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  6. #6
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    You seem to have solved for Z + 1/Z and not Z + 1/sqrt Z
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    Given that z=-2+i , express  z+\frac{1}{\sqrt{z}} in the form of a+bi
    <br />
\sqrt z  = \sqrt { - 2 + i}  = a + bi,<br />

    <br />
\therefore  - 2 + i = (a + bi)^2  = a^2  + 2abi - b^2 ........\langle 1\rangle <br />
    <br />
a^2  - b^2  =  - 2,2ab = 1 \Rightarrow b^2  = \frac{1}<br />
{{4a^2 }}<br />

    <br />
\text{Substitute into }\langle 1\rangle \text{: }4a^4  + 8a^2  - 1 = 4(a^2 )^2  + 8a^2  - 1\text{ }.........\langle \text{2}\rangle <br />

    <br />
\text{Put }a^2  = x \Rightarrow \langle \text{2}\rangle  = 4x^2  + 8x - 1<br />
    <br />
\text{Set }\langle \text{2}\rangle  = 0\text{ and solve: }a^2  =  - 1 + \frac{1}<br />
{2}\sqrt 5  \Rightarrow a = \sqrt { - 1 + \frac{1}<br />
{2}\sqrt 5 } \text{    (only take + ve part)}<br />
    <br />
\text{From }2ab = 1\text{, }b = \frac{1}<br />
{{2a}} = \frac{1}<br />
{2}\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{ - \tfrac{1}<br />
{2}} <br />

    <br />
\therefore z =  - 2 + i = (a + bi)^2  = <br />

    <br />
 = \left[ {\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{\tfrac{1}<br />
{2}}  + \frac{1}<br />
{2}\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{ - \tfrac{1}<br />
{2}} i} \right]^2 <br />

    <br />
\text{and  }z + \frac{1}<br />
{{\sqrt z }} = <br />
    <br />
 = \left[ {\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{\tfrac{1}<br />
{2}}  + \frac{1}<br />
{2}\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{ - \tfrac{1}<br />
{2}} i} \right]^2  + <br />
    <br />
 + \left[ {\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{\tfrac{1}<br />
{2}}  + \frac{1}<br />
{2}\left( { - 1 + \frac{1}<br />
{2}\sqrt 5 } \right)^{ - \tfrac{1}<br />
{2}} i} \right]^{ - 1} <br />

    <br />
\text{which we need to reduce in the form of }a + bi.\text{ Any hints are highly appreciated}\text{.}
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