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Given $z= \dfrac{i-1}{2}$. Evaluate $z^{1/2}$ and show the roots.

I got $z^{1/2}=\dfrac{1}{2^{1/4}}\left(\cos\dfrac{3\pi}{8} + i\sin\dfrac{3\pi}{8}\right)$

First of all is my $z^{1/2}$ correct? And secondly how to show the roots??

I would be glad and appreciated if someone could help me out.

David
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3 Answers3

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Partly correct; what you have found is the principle value. A complete answer should be: $$z^{1/2}=\left(\frac{1}{\sqrt{2}}e^{i\frac{3\pi}{4}+2n\pi}\right)^{1/2}=\frac{1}{2^{1/4}}e^{i\frac{3\pi}{8}+n\pi}=\frac{1}{2^{1/4}}\left(\cos\left(\frac{3\pi}{8}+n\pi\right)+i\sin\left(\frac{3\pi}{8}+n\pi\right)\right)=\pm\frac{1}{2^{1/4}}\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}\right)$$

  • Hi Julien Godawatta. Thank you for helping me. You showed me another variety of solving it using exponentials. Thank you again Julien – David Sep 24 '12 at 18:44
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$z=-\frac12+\frac i 2=R(\cos y+i\sin y)(say)$ where $R>0$

$R\cos y=-\frac12, R\sin y=\frac12\implies R^2=(-\frac12)^2+(\frac12)^2=\frac12$

So, $\tan y=\frac{R\sin y}{R\cos y}=-1\implies y=2n\pi+\frac{3\pi}{4}$ as $\cos y<0$ and $\sin y>0$ ,so $y$ lies in the 2nd quadrant.

So, $z =\frac 1{\sqrt2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$

So, the general value of $z^{\frac12}$ is $(\frac12)^{\frac14}(\cos(n\pi+\frac{3\pi}{8})+i\sin(n\pi+\frac{3\pi}{8}))$ where $n=0,1$ using de Moivre's formula.

$n=0\implies z^{\frac12}=(\frac12)^{\frac14}(\cos(\frac{3\pi}{8})+i\sin(\frac{3\pi}{8}))$

$n=1\implies z^{\frac12}=(\frac12)^{\frac14}(\cos(\pi+\frac{3\pi}{8})+i\sin(\pi+\frac{3\pi}{8}))=-(\frac12)^{\frac14}(\cos(\frac{3\pi}{8})+i\sin(\frac{3\pi}{8}))$

One may look into this, for the values of $n$.

Community
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lab bhattacharjee
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  • Thank you so much lab bhattacharjee. You really helped me well. Thanks for your effort man!!! – David Sep 24 '12 at 18:43
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For computing $z^{1/2}=\sqrt{x+yi}$ put $$\sqrt{x+yi}=a+bi$$ and so $(\sqrt{x+yi})^2=(a+bi)^2$. If you do the latter identity, you will find $$(1): a^2-b^2=x\\(2):2ab=y\\(3):(a^2+b^2)^2=\sqrt{x^2+y^2}$$ Adding (1) and (3) gives you $$2a^2=\sqrt{x^2+y^2}+x$$ and subtracting (1) of (3) gives you $$2b^2=\sqrt{x^2+y^2}-x$$ Now, the rest is easy.

Mikasa
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