Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
See here. – Mhenni Benghorbal Sep 24 '12 at 18:23
@David: Have a look at my answer as well. – Babak S. Sep 24 '12 at 19:16
up vote 1 down vote accepted

$z=-\frac12+\frac i 2=R(\cos y+i\sin y)(say)$ 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 $n$.

share|cite|improve this answer
Thank you so much lab bhattacharjee. You really helped me well. Thanks for your effort man!!! – David Sep 24 '12 at 18:43

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

share|cite|improve this answer
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

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.

share|cite|improve this answer
+1 for my friend, Babak! – amWhy Mar 22 '13 at 0:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.