these are my toughs:
$$z^2 = 1 + 2i \Longrightarrow (x+yi)(x+yi) = 1 + 2i$$
so: $x^2-y^2 = 1$ and $2xy = 2$
then i got that $x = 1/y$ but i cant continue to find the real- and imaginary part of z anymore. Appriciated any help
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1Plug that $x=1/y$ into your first equation to get a quadratic in $y^2$. – Zain Patel May 28 '17 at 13:15
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Here's a MathJax tutorial :) – Shaun May 28 '17 at 13:15
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yeah then i get 1/y^2 minus y^2 but this isnt helpful either – Razmo May 28 '17 at 13:15
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4Yes it is. Multiply throughout by $y^2$ and you'll get a quadratic in $y^2$. – Zain Patel May 28 '17 at 13:16
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2An important principle of mathematics is: Don't Stop. – Lee Mosher May 28 '17 at 13:17
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well i got z^2 = (1/y^2) - y^2 + 2i – Razmo May 28 '17 at 13:19
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A simple way to get one more equation : use $\left|z^2\right|=\left|1+2i\right|$, so $x^2+y^2=\sqrt5$. Combine with $x^2-y^2=1$. $2xy=2$ then gives you the choice of signs for the roots. – Nicolas FRANCOIS May 28 '17 at 13:20
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1Possible duplicate of How do I get the square root of a complex number? – Hans Lundmark May 28 '17 at 13:21
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How...? You have $x^2 - y^2 = 1$. So $x^2y^2 - y^4 = y^2$, which gives $1 - y^4 = y^2$ or equivalently $1 - u^2 = u$ where $u = y^2$. Now solve this quadratic in $u$. – Zain Patel May 28 '17 at 13:21
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@ZainPatel ok i got y^2 = 1-y^6 or y = sqrt(1-y^6). What now? – Razmo May 28 '17 at 13:26
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@Razmo see my answer. I'm not sure how you managed to get that. – Zain Patel May 28 '17 at 13:30
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i didnt quite understand what you did mean – Razmo May 28 '17 at 13:31
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In general, we have this
Lemma: If $z=a+ib \in \mathbb{C}$, with $a,b\in\mathbb{R}$, then $$ w = \sqrt{\frac{|z|+a}{2}} + i\epsilon\sqrt{\frac{|z|-a}{2}},$$ where $\epsilon =\pm 1$ according to $b=\epsilon|b|$, satisfies $w^2 = z$.
Proof: Let $w=x+iy$ satisfying $w^2=z$. Then $x^2 - y^2+2xyi =a+bi$. This equation is equivalent to the system $$ x^2 - y^2 = a \text{ ; } 2xy = b.$$ Since $w^2 = z$, we have $x^2+y^2 = |z|$ too, and we can conclude that $$x^2 = \frac{|z|+a}{2} \text{ ; } y^2 = \frac{|z|-a}{2}.$$ Choosing the positive square roots, we can write $$x = \sqrt{\frac{|z|+a}{2}} \text{ ; } y = \epsilon\sqrt{\frac{|z|-a}{2}}$$ satisfying $2xy = b$, i.e., $\epsilon = 1$ if $b > 0$ and $\epsilon = -1$ if $b < 0$.
Rodrigo Dias
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You have $x^2 - y^2 = 1$ and $x = \frac{1}{y}$. Substitute the latter into the former to get $$\frac{1}{y^2} - y^2 = 1 \implies 1 - y^4 = y^2 \implies y^4 + y^2 - 1 =0.$$
You can solve the quadratic $u^2 + u -1 = 0$ giving solution $u = \frac{1}{2}(-1 \pm \sqrt{5})$. So you need to solve $y^2 = \frac{1}{2}(-1 + \sqrt{5})$ (since $y$ is real) which is now straightforward.
Zain Patel
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Do you now how to solve a quadratic $u^2 + u - 1 = 0$? Using the quadratic formula? – Zain Patel May 28 '17 at 13:39
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Not sure what you mean. We have $u = \frac{-1 \pm \sqrt{1^2 +4}}{2}$ – Zain Patel May 28 '17 at 13:42
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but how can u go from this construc y^4+y^2-1 = 0 to the quadratic one with u^2+u-1=0? – Razmo May 28 '17 at 13:46
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@Razmo no because $y$ is real and so its square must be non-negative. – Zain Patel May 28 '17 at 14:13
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An alternative approach would be to use DeMoivre's Theorem. $1+2i=\sqrt{5}e^{i\arctan{2}}$ so $z=\sqrt[4]{5}e^{i\arctan 2/2}$ or $z=\sqrt[4]{5} e^{(i\arctan 2 +2\pi i)/2}$ If $\tan \theta =2, \tan \frac{\theta}{2} = \frac{2}{1+\sqrt{5}}$ giving $$z=\sqrt[4]{5}\frac{1+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}+i\sqrt[4]{5}\frac{2}{\sqrt{10+2\sqrt{5}}}$$ The other root being $-z$.
sharding4
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\begin{align} z^2 = 1 + 2i & = |1+2i|(\cos\alpha+i\sin\alpha) \text{ where } \tan \alpha = \frac 2 1 \\[10pt] & = \sqrt{1^2+2^2} (\cos\alpha+i\sin\alpha) = \sqrt 5 (\cos\alpha+i\sin\alpha) \end{align} Therefore $$ z = \pm\left( \sqrt{\sqrt 5} \right)\left( \cos\frac\alpha2 + i \sin\frac\alpha 2 \right). $$