2
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for some reason I can to figure out how to rewrite this square root. I have:

$\sqrt{2+i}$

And I need to rewrite it into:

$\frac{\sqrt{2(\sqrt{5} + 2)} + \sqrt{-2(\sqrt{5} - 2)}}{2}$

Can anybody show me I to do this? I've been trying for an hour now..

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  • 6
    $\begingroup$ Hint: $\sqrt{\cos \theta + i \sin \theta} = \cos(\theta/2) + i \sin(\theta / 2)$. Note $2 + i = \sqrt{5} \cdot \left(\frac{2}{\sqrt 5} + i \frac{1}{\sqrt 5}\right)$ $\endgroup$ – Omnomnomnom Jun 27 '14 at 19:44
  • $\begingroup$ Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$. $\endgroup$ – Yves Daoust Jun 27 '14 at 19:54
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    $\begingroup$ @DaneBouchie Read that again. I say $$ 2 + i = \sqrt{5} \cdot \left(\frac{2}{\sqrt 5} + i \frac{1}{\sqrt 5}\right) $$ $\endgroup$ – Omnomnomnom Jun 27 '14 at 20:18
  • $\begingroup$ Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf. $\endgroup$ – Dane Bouchie Jun 27 '14 at 20:22
2
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One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic. $$(x+yi)^2=2+i$$ so

$$x^2-y^2=2$$ $$2xy=1$$

and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$

So $$x^2+y^2=\sqrt{5}$$ (positive since $x$ and $y$ real) and $$x^2=\frac{\sqrt{5}+2}{2}$$ $$y^2=\frac{\sqrt{5}-2}{2}$$

so

$$x=\pm\sqrt{\frac{\sqrt{5}+2}{2}}$$ $$y=\pm\sqrt{\frac{\sqrt{5}-2}{2}}$$

Both $x$ and $y$ have same sign since $2xy=1$.

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