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I am stuck on the problem as the solutions given at two places do not either concur or or not understandable by me.

The non concurring answer (here, pg #12) states: one root is $-\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}$.

I feel my approach will yield as follows:

$x^2 +i = 0 \implies x = (-i)^{1/2} \implies$ modulus = $1$, and by de-Moivre's formula with angle $= -\pi/2$, i.e. $-i = 1.(\cos(-\pi/2) + i.\sin(-\pi/2))$.
=> $x_1 = 1.(\cos(-\pi/4) + i.\sin(-\pi/4))$

But, in fourth quadrant the cosine is positive, so one (particular) root is : $\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}$, and the other is diametrically opposite :$-\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$, that is found by adding $\pi (= n2\pi/m = n^{'}\pi/m)$ to the original answer, i.e. $x_2 = 1.(\cos(3\pi/4) + i.\sin(3\pi/4)) \implies x_2 = -\frac{1}{\sqrt2} + i\frac{1}{\sqrt2}$.

The not understandable answer is given at wolframalpha that although gels with me in terms of geometric location of the points, yet the answer of $\pm(-1)^{3/4}$ confuses me totally.

jiten
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  • The wolframalpha solution is the same as yours – John Doe Jan 11 '18 at 02:49
  • @JohnDoe But not able to see the expression form to be so. – jiten Jan 11 '18 at 02:50
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    $(-1)^{3/2} = -(-1)^{1/2} = -i$. Therefore $(-1)^{3/4} = (-i)^{1/2}$ – Dylan Jan 11 '18 at 02:51
  • @Dylan Thanks for that. – jiten Jan 11 '18 at 02:54
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    @jiten It doesn't tell you much in that it isn't in polar form, and isn't written as $x+iy$. But it is correct. If you define the branch of $f(z)=z^{\frac14}$ to be $z^{\frac14}=e^{\frac14 \log(|z|)+\frac i4\arg(z)}$, then $$(-1)^\frac34=(e^{\frac i4\cdot \pi})^3=e^{\frac34i\pi}=-\frac1{\sqrt{2}}+i\frac1{\sqrt2}$$ as required. (although I suspect this is more advanced than you're expected to do for your work) – John Doe Jan 11 '18 at 03:01
  • @JohnDoe Thanks a lot for giving a better view. If possible, kindly state in detail or please give some reference terms to google for regarding the "branch of $z^\frac{1}{4}$". Else, please refer to some source. – jiten Jan 11 '18 at 03:28
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    Are you aware of branch cuts? The nth root functions has n branches for each root – Dylan Jan 11 '18 at 03:38
  • @Dylan Read this as cursory in complex analysis. So, these are linked topics! Thanks for that. – jiten Jan 11 '18 at 03:39

2 Answers2

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Your geometric reasoning is completely correct: the roots must be in the second and fourth quadrants.

The answer $-\frac{1}{\sqrt2} -i\frac{1}{\sqrt2}$ must therefore be wrong. We can also check that it is wrong by multiplying it out: $(-\frac{1}{\sqrt2} -i\frac{1}{\sqrt2})(-\frac{1}{\sqrt2} -i\frac{1}{\sqrt2}) = \frac{1}{2} + 2i\frac{1}{2} - \frac{1}{2} = i \neq -i$. On the other hand, multiplying out your two answers will indeed give $-i$.

$\pm(-1)^\frac{3}{4}$ works too, since $(\pm(-1)^\frac{3}{4})^2 = (-1)^\frac{3}{2} = ((-1)^\frac{1}{2})^3 = i^3 = -i$. However, it's not as enlightening in terms of where on the complex plane those points actually are.

BallBoy
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  • The books states it as 'one of the roots', so the other must be having both real & imaginary parts positive, leading to the same value ($-i$), so that's not the criteria. But, it must a good idea, may be in some other situation for cross-checking, but not valid here. – jiten Jan 11 '18 at 02:55
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    @jiten Your link isn't working for me, so I haven't seen the book... By "both real & imaginary parts are positive," I assume you mean $\frac{1}{\sqrt2} + i\frac{1}{\sqrt2}$. But if you square this, you also get $i$, not $-i$... – BallBoy Jan 11 '18 at 02:57
  • I hoped that both roots need to be multiplied, sorry as my logic is wrong. Actually, never thought like that before, and your comment made it clear that you are correct. – jiten Jan 11 '18 at 02:59
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Alternative Method: Let $z=x+iy$ and $c=a+ib$. Consider the equation $z^2=c$. which is $x^2-y^2+2ixy=a+ib$. Solve for $x$ and $y$. We get the desired result.

Here, $x^2-y^2=0,2xy=-1.\implies y=-x, 2x^2=1$ Hence $z=\frac{1}{\sqrt 2}-i \frac{1}{\sqrt 2}$ and $z= -\frac{1}{\sqrt 2}+i \frac{1}{\sqrt 2}$