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Solve in $\mathbb{C}$ the following equatios:

$$z^2=\frac{1}{2}-i\frac{\sqrt{2}}{2}$$

I was solving the exercise from the book A to Z complex numbers, I tried this in two ways one by taking $z=x+iy$ but then I am getting another quadratic equation in $y$ where the constant term is a complex number. I also tried to get it in the form $re^{i\theta}$ but there $\theta=\arctan{\sqrt{2}}$. Any help will be truly appreciated.

Sebastiano
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Sayantan
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    "I am getting another quadratic equation" No, you should be getting two equations with real coefficients, no imaginary units involved. Why don't you edit the question and show this step. – dxiv May 29 '21 at 21:48
  • You can compute $\cos(\arctan(t)/2)$ and $\sin(\arctan(t)/2)$ in terms of $t,$ up to sign. https://colalg.math.csusb.edu/~devel/IT/main/m05_identities/src/s03_half-angles.html – Thomas Andrews May 29 '21 at 21:54

4 Answers4

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Notice that assuming $z=x+yi$ you gain the advantage of looking at real numbers only. Now $x^2-y^2=\frac{1}{2}$ and $2xy=-\frac{\sqrt{2}}{2}$ can be solved in reals by substitution. (You would get more than 2 answers, but only two answers are in reals)

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Hint

Check the corolary to Moivre's Theorem:

The n-th roots of a complex number $z=r\cdot e^{i\theta}=r\cdot(\cos\theta +i\sin\theta)$ are:

$z_m=r^{\frac{1}{n}}\cdot e^{i\left(\frac{\theta}{n}+\frac{2m\pi}{n}\right)}=r^{\frac{1}{n}}\cdot(\cos\left(\frac{\theta}{n}+\frac{2m\pi}{n}\right) +i\sin\left(\frac{\theta}{n}+\frac{2m\pi}{n}\right))$ for $m=0, ..., n-1$.

GoRza
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The easy way

$|z|^2 = \frac {1}{2}^2 + (\frac {\sqrt 2}{2})^2 = \frac {3}{4}\\ |z| = \frac {\sqrt 3}{2}\\ z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}} e^{\frac {i}{2}\arctan \sqrt 2}$

You really could stop here. Or, you can keep going:

$z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}}(\cos(\frac {1}{2}\arctan \sqrt 2) + i\sin(\frac {1}{2}\arctan \sqrt 2))\\ z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}}\left(\sqrt {\frac {\sqrt 3+ 1}{2\sqrt 3}} + i\sqrt {\frac {\sqrt 3- 1}{2\sqrt 3}}\right)\\ z^{\frac 12} = \frac{\sqrt {\sqrt 3+ 1}}{2} + i\frac{\sqrt {\sqrt 3- 1}}{2}$

And we can compare this to our other approach

$z^{\frac 12} = x+iy\\ (x+iy)^2 = \frac {1}+i\frac {\sqrt 2}{2}\\ x^2 - y^2 = \frac 12\\ 2xy = \frac {\sqrt 2}{2}\\ y = \frac {\sqrt 2}{4x}\\ x^2 - \frac {1}{8 x^2} = \frac 12\\ 8x^4 - 4x^2 - 1 = 0\\ x^2 = \frac {4 + \sqrt {48}}{16}\\ x^2 = \frac {1 + \sqrt {3}}{4}\\ x = \frac {\sqrt {1 + \sqrt {3}}}{2}\\ y = \frac {\sqrt {-1 + \sqrt {3}}}{2}\\ z^{\frac 12}= \frac {\sqrt {1 + \sqrt {3}}}{2} + i\frac {\sqrt {-1 + \sqrt {3}}}{2}$

Doug M
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Hints for the high school way:

Set $z=x+iy$. Identifying the real and imaginary parts, we obtain the equations $$\begin{cases} x^2-y^2=\frac12\\[1ex] 2xy=-\frac{\sqrt{2^{\vphantom{b}}}}2 \end{cases}$$ We also know that $\:|z|^2=x^2+y^2=\biggl|\frac12-i\frac{\sqrt 2}2\biggr|=\frac{\sqrt 3}2$, so that we have a linear system in $x^2$ and $y^2$: $$\begin{cases} x^2-y^2=\frac12\\[1ex] x^2+y^2=\frac{\sqrt{3^{\vphantom{b}}}}2 \end{cases}\iff x^2= \frac{1+\sqrt 3}4,\;y^2=\frac{-1+\sqrt 3}4$$ and we obtain two opposite solutions for $x$ and for $y$. However, the identification of the imaginary parts shows that $x$ and $y$ have opposite signs and ultimately the there are two values for $z$.

Bernard
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