# Complex numbers - finding a square root of something

Let $z_1 , z_2$ be two complex numbers that satisfy:

$\dfrac{z_2 } {\bar{z_1}}= \frac{3}{8} \big(\cos(75^{\circ})+i\sin(75^{\circ})\big)$ ,

$z_1 z_2 ^2 = \frac{1}{3} \big(\cos(120^{\circ}) + i\sin(120^{\circ}) \big)$ .

How can I determine with of the following can be a possible value for $\sqrt{z_1}$ ?

(a) $\frac{2}{\sqrt{3}} \mbox{cis}(135^{\circ})$

(b) $\frac{2}{3} \mbox{cis}(155^{\circ})$

(c) $\frac{2}{\sqrt{3}} \mbox{cis}(195^{\circ})$

(d) $\frac{2}{\sqrt{3}} \mbox{cis}(175^{\circ})$

(e) $\frac{2}{3} \mbox{cis}(215^{\circ})$

thanks !!!

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Those should probably all be \cos, right? –  icurays1 Nov 30 '12 at 17:54
$\mathop{\text{cis}}$ is a standard, but uncommon, notation for the function $x \mapsto \cos(x) + \mathbf{i} \sin(x)$ –  Hurkyl Nov 30 '12 at 17:55
@yuta, how $z$ is linked with $z_1$ or $z_2$? –  lab bhattacharjee Nov 30 '12 at 17:56
I meant $\sqrt{z_1}$ . Sorry! –  yuta Nov 30 '12 at 17:58

Write the $z_i$ in polar representation as $z_1 = r_1 \mathrm{cis}(\theta_1)$ and $z_2 = r_2 \mathrm{cis}(\theta_2)$. Then, the two equations you have are: $$\frac{z_2}{\bar{z_1}} = (r_2 \mathrm{cis}(\theta_2))(\frac{1}{r_1} \mathrm{cis}(\theta_1)) = \frac{r_2}{r_1}\mathrm{cis}(\theta_1 + \theta_2) = \frac{3}{8}\mathrm{cis}(75^{\circ}),$$ $$z_1 z_2^2 = r_1 r_2^2 \mathrm{cis}(\theta_1 + 2\theta_2) = \frac{1}{3} \mathrm{cis}(120^{\circ}).$$

What does the equations imply on $r_i$ and $\theta_i$?

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thanks a lot ! !!!!!!! –  yuta Dec 1 '12 at 8:11

If you have: z1 = a*[$\cos (x_1)$ + i*$\sin (x_1)$] and z2 = b*[$\cos (x_2)$ + i*$\sin (x_2)$], then you have:

$z_2/z_1$ = a/b * [$\cos (x_2-x_1)$ + i*$\sin (x_2-x_1)$] , and from this equation and your first one you`ll have:

$a/b$ = 3/8 and x2-x1= 75.

$z_2^2$ = $b^2$ [$\cos (2x_2)$ + i*$\sin (2x_2)$] (Moivre) and you have:

$z_1*z_2^2$= $a*b^2$ *[$\cos (x_1+2x_2^2)$ + i*$\sin (x1+2x2^2)$]; so from the second equation result that :

$a*b^2$ = 1/3;

x1+ 2*x2= 120;

With a/b = 3/8 and x2-x1= 75.

We have 3*x2=195 so x2= 65, then x1 = -10 , and a= 4/3

So x1 = 4/3 [$\cos (350)$ + i*$\sin (350)$]

For the square root of x you have:

So: we hace $x_3$ = c * [$\cos (x)$ + i*$\sin (x)$] as the root of x1

From the formula we have c= $2/\sqrt3$. and x= 350/2 =175.

$x_3$ = $2/\sqrt3$ * [$\cos (175)$ + i*$\sin (175)$]

Final result D

***Sorry for the writing, I am new , and I am now learning haow to write mathematics symbol in stackexchange!

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