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Square root of complex number

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    show that [tex]\sqrt{1+ja}[/tex] is equivalent to [tex]\pm(1+j)(a/2)^{1/2}[/tex] with a>>1

    2. Relevant equations
    Euler's formula?

    3. The attempt at a solution
    with a>>1
    |z| = [tex]\sqrt{(1 + a^{2})}[/tex] == a
    lim a-->infinity arctan (a/1) == [tex]\pi/2[/tex]
    [tex]\sqrt{z}[/tex] = [tex]\sqrt{(ae^{j\pi/2})}[/tex]
    [tex]\sqrt{z}[/tex] = [tex]\pmsqrt a^{1/2} e^{j1.25}[/tex]

    However, when I transfer back to complex form, I don't get it to equal 1+j. Not too sure how they got a/2 as well.

    Any tips would be great.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 15, 2010 #2


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    Homework Helper

    Where did you get 1.25 from, in the last line? That looks wrong...

    I actually wouldn't bother with converting to decimals at all, since it would be inexact. Just leave it as [itex]e^{j\pi/2}[/itex]. What is the square root of that number?
  4. Jan 15, 2010 #3
    That came from the square root of pi / 2.
  5. Jan 15, 2010 #4


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    Staff Emeritus
    Science Advisor

    You can't, it's not true. [itex][(1+ j)(a/2)^{1/2}]^2= (1+j)^2 (a/2)= (1+ 2j- 1)(a/2)= aj[/itex], not 1+ja. Of course, if, by "a>>1" you mean a is extremely large, then a is large compared to 1 so that is approximately true. Is that what you meant by "equivalent"?

  6. Jan 15, 2010 #5


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    Homework Helper

    There was no reason to take the square root of [itex]\pi/2[/itex]. Do you remember how square roots act on exponentials? That is, in general, what's the square root of [itex]e^x[/itex]?

    HallsofIvy makes a good point, too - I didn't realize that at first.
  7. Jan 15, 2010 #6
    Yeah I probably should have said approximately instead of equivalent. The question I posted is word for word out of an applied electromagnetism book for engineers. When it says assume a>>1, I just assumed a to be infinity.
  8. Jan 15, 2010 #7
    WHOOPS, I meant to type 0.785 instead of 1.25. I guess thats what I get for trying to simplify without looking at my hand written work. In either case, it still doesn't seem to work out. I'm kind of stumped as how to prove the (a/2)^1/2 part.
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