roots of complex numbers $z^2$

Question

Let $z^2 = \frac{1}{2} + \frac{\sqrt{3}}{2i}$

Where $z$ is an element of the Complex Numbers

Find the two possible values of $z$

I would really appreciate if someone could help me with this question. Thanks

Answer 1

Just take the square root of this.

It's easiest if you realize this is a nice number on the unit circle (recognize the cosine and sine of $60^\circ$),

$$z^2=e^{i\pi/3+2k\pi i}$$ where I took into account the periodicity of the exponential notation. Now just

$$z=e^{(i\pi/3+2k\pi i)/2}=e^{i\pi/6+k\pi i}$$ which yields two different results for k=0 and k=1 before it starts repeating again.

Now just rewrite it back into cartesian notation with the Euler's formula: $$z_1=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}$$ $$z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$$ You can continue yourself.

p.s. if your $/2i$ means division by $i$ instead of multiplication, just invert the sign in the exponent to $e^{-i\pi/3+2k\pi i}$ and conjugate the results.

Answer 2

Hint: the purely Cartesian approach would be to let $z = a + bi$ for some real numbers $a,b$. Then, we have

$$z^2 = (a^2 - b^2) + 2abi$$

But $z^2 = \frac{1}{2} + \frac{\sqrt{3}}{2}i$. So we equate the real parts, and the imaginary parts, and solve simultaneously to find the possible values of $a,b$.

Answer 3

Both this (in the edited version) and the original are points on the unit circle, thus of form $\cos\theta+i\sin\theta$. According to DeMoivre, all you need to do is bisect the angle $\theta$ — two values, of course, differing by $\pi$.

Answer 4

Denesting $\,\sqrt{a+b\sqrt{n}}\,$ can be done by a simple rule that I discovered in my youth.

Simple Denesting Rule $\rm\ \ \, \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

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Here $\:1/2-i\sqrt{3}/2\:$ has norm $= 1.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\ {-}1/2-i\sqrt{3}/2\:$

which has ${\rm\ \sqrt{trace}}\: =\: \sqrt{-1}\ =\ i.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\ \ \ i/2-\sqrt{3}/2$

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Or: $\:1/2-i\sqrt{3}/2\:$ has norm $= 1.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = \ {-}1 \ $ yields $\ 3/2-i\sqrt{3}/2\:$

which has $\rm\ \sqrt{trace}\: =\: \sqrt{3}.\ $ Then $\rm\,\ \color{brown}{dividing\ it\ out}\ $ of the above yields $\ \ \ \ \sqrt{3}/2-i/2$

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One may choose the sign of the square-root so that the arithmetic is simpler, as in the first case above. For many further worked examples see my prior posts on denesting.