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Daniel za post:
stevan95
Reputacija: 4.55%
od Daniel » Ponedeljak, 01. Jul 2013, 19:01
[dispmath]z^4=-16[/dispmath][dispmath]z=x+iy,\quad x,y\in\mathrm{R}[/dispmath][dispmath]\left(x+iy\right)^4=-16[/dispmath][dispmath]\left(x^2-y^2+i2xy\right)^2=-16[/dispmath][dispmath]\left(x^2-y^2\right)^2-4x^2y^2+i4xy\left(x^2-y^2\right)=-16[/dispmath]
Pošto je izraz na desnoj strani realan, mora biti realan i izraz na levoj strani, tj. imaginaran deo mu mora biti jednak nuli:
[dispmath]i4xy\left(x^2-y^2\right)=0[/dispmath][dispmath]x=0\quad\lor\quad y=0\quad\lor\quad x^2=y^2[/dispmath]
[inlmath]I\quad x=0:[/inlmath]
[dispmath]\left(iy\right)^4=-16[/dispmath][dispmath]y^4=-16[/dispmath]
Pošto [inlmath]y\in\mathrm{R}[/inlmath], ovaj slučaj nema rešenja.
[inlmath]II\quad y=0:[/inlmath]
[dispmath]x^4=-16[/dispmath]
Pošto [inlmath]x\in\mathrm{R}[/inlmath], ni ovaj slučaj nema rešenja.
[inlmath]III\quad x^2=y^2:[/inlmath]
[dispmath]\cancelto{0}{\left(x^2-y^2\right)^2}-4x^2y^2+\cancelto{0}{i4xy\left(x^2-y^2\right)}=-16[/dispmath][dispmath]-4x^2y^2=-16[/dispmath][dispmath]x^2y^2=4[/dispmath][dispmath]xy=\pm 2[/dispmath]
[inlmath]x^2=y^2\quad\Rightarrow\quad x=\pm y[/inlmath]
[inlmath]x=\pm y\;\land\;xy=\pm 2[/inlmath]
[inlmath]\Rightarrow\quad x=-\sqrt 2,\;y=-\sqrt 2\;\lor\;x=-\sqrt 2,\;y=\sqrt 2\;\lor\;x=\sqrt 2,\;y=-\sqrt 2\;\lor\;x=\sqrt 2,\;y=\sqrt 2[/inlmath]
Ukupan skup rešenja:
[dispmath]z\in\left\{-\sqrt 2-i\sqrt 2,\;-\sqrt 2+i\sqrt 2,\;\sqrt 2-i\sqrt 2,\;\sqrt 2+i\sqrt 2\right\}[/dispmath]
Drugi način: preko trigonometrijskog zapisa kompleksnog broja
[dispmath]z=\sqrt[4]{-16}[/dispmath][dispmath]z=\sqrt[4]{16\left[\cos\left(2k+1\right)\pi+i\sin \left(2k+1\right)\right]}=\sqrt[4]{16e^{i\left(2k+1\right)\pi}}=[/dispmath][dispmath]=\sqrt[4]{16}e^{i\frac{\left(2k+1\right)\pi}{4}}=2e^{i\left(\frac{\pi}{4}+\frac{k\pi}{2}\right)}=2\left[\cos\left(\frac{\pi}{4}+\frac{k\pi}{2}\right)+i\sin\left(\frac{\pi}{4}+\frac{k\pi}{2}\right)\right][/dispmath][dispmath]k\in\mathrm{Z}[/dispmath][dispmath]k=4n:\quad z=2\left[\cos\left(\frac{\pi}{4}+2n\pi\right)+i\sin\left(\frac{\pi}{4}+2n\pi\right)\right]=2\left(\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)=\sqrt 2+i\sqrt 2[/dispmath][dispmath]k=4n+1:\quad z=2\left[\cos\left(\frac{3\pi}{4}+2n\pi\right)+i\sin\left(\frac{3\pi}{4}+2n\pi\right)\right]=2\left(\frac{-\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)=-\sqrt 2+i\sqrt 2[/dispmath][dispmath]k=4n+2:\quad z=2\left[\cos\left(\frac{5\pi}{4}+2n\pi\right)+i\sin\left(\frac{5\pi}{4}+2n\pi\right)\right]=2\left(\frac{-\sqrt 2}{2}-i\frac{\sqrt 2}{2}\right)=-\sqrt 2-i\sqrt 2[/dispmath][dispmath]k=4n+3:\quad z=2\left[\cos\left(\frac{7\pi}{4}+2n\pi\right)+i\sin\left(\frac{7\pi}{4}+2n\pi\right)\right]=2\left(\frac{\sqrt 2}{2}-i\frac{\sqrt 2}{2}\right)=\sqrt 2-i\sqrt 2[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain