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Ahsome

  • one year ago

Absolute Value question

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  1. ahsome
    • one year ago
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    Assuming \(|x+iy-1| = |x+iy-2i|\), express \(y\) in terms of \(x\) Where \(x,y\in\mathbb{R}\)

  2. ahsome
    • one year ago
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    Any idea, @ganeshie8?

  3. anonymous
    • one year ago
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    |dw:1433247964097:dw|

  4. anonymous
    • one year ago
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    z is a complex number equidistant from (1,0) and (0,2) so it is a straight line perpendicular to the line joining both of them and passing through the mid point you will get the relation between y and x

  5. ganeshie8
    • one year ago
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    Nice!

  6. ahsome
    • one year ago
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    I am confused. What do you mean by \(\mathbf{equidistant }\)?

  7. ganeshie8
    • one year ago
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    let \(z=x+iy\) then the given equation is same as \[|z-1| = |z-2i|\]

  8. ganeshie8
    • one year ago
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    recall that |z-1| gives you the distance between \(z\) and \(1+0i\) and |z-2i| gives the distance between \(z\) and \(0+2i\)

  9. ganeshie8
    • one year ago
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    if you don't like that geometric thingy, you may work it algebraically : \[|x+iy-1| = |x+iy-2i|\\~\\|(x-1)+iy|=|x+i(y-2)|\\~\\(x+1)^2+y^2 = x^2+(y-2)^2\] simplify

  10. ahsome
    • one year ago
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    May I ask, why you squared everything?

  11. ahsome
    • one year ago
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    NVM, REALISED

  12. ahsome
    • one year ago
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    I get \[y=\dfrac{2x}{4}+\dfrac{3}{4}\]. Is that right?

  13. ahsome
    • one year ago
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    Wait, no

  14. ahsome
    • one year ago
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    Why did: \[|(x−1)+iy|=|x+i(y−2)| \]Become\[ (x+1)^2+y^2=x^2+(y−2)^2\], Instead of\[ (x-1)^2+y^2=x^2+(y−2)^2\]

  15. ganeshie8
    • one year ago
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    Ahh thats just my mistake!

  16. ahsome
    • one year ago
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    Ahh, phew ;)

  17. ahsome
    • one year ago
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    The answer I get is \[\huge{y=\dfrac{x}{2}+\dfrac{3}{4}}\]

  18. ahsome
    • one year ago
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    The answer in the book is\[\huge{y=\dfrac{x}{4}+\dfrac{3}{4}}\]

  19. ahsome
    • one year ago
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    What do you think is the issue, @ganeshie8?

  20. ganeshie8
    • one year ago
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    y = x/2 + 3/4 is right

  21. ganeshie8
    • one year ago
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    your textbook must be having a typo

  22. ahsome
    • one year ago
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    Ahh, ok then. Thanks so much!. Could you help me with another question?

  23. ganeshie8
    • one year ago
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    wil try, pls post as a new q so that it gets bumped in the top and others see

  24. ahsome
    • one year ago
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    Ok then :D

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