Witam,
jak sprowadzić liczbę zespoloną w postaci \(\displaystyle{ \frac{4-\sqrt{48}i}{-\sqrt{2}+ \sqrt{2}i}}\) do postaci \(\displaystyle{ -2\cdot2^{\frac{3}{4}}(1-i \sqrt{3})^{\frac{1}{4}}}\) ?
Zaczynam od pomnożenia ułamka przez sprzężenie lecz kompletnie nie mam pomysłu co dalej
Sprowadzenie liczby zespolonej do innej postaci.
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bb314
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Sprowadzenie liczby zespolonej do innej postaci.
\(\displaystyle{ \blue a=\frac{4-\sqrt{48}i}{-\sqrt{2}+ \sqrt{2}i}\black=\frac{\left(4-\sqrt{48}i\right)\left(-\sqrt2-\sqrt2i\right)}{\left(-\sqrt{2}+ \sqrt{2}i}\right)\left(-\sqrt2-\sqrt2i\right)}=-\sqrt6-\sqrt2+(\sqrt6-\sqrt2)i}\)
\(\displaystyle{ |a|=\sqrt{\left( -\sqrt6-\sqrt2\right)^2+\left( \sqrt6-\sqrt2\right)^2}=\sqrt{\left(6+2\sqrt6\sqrt2+2\right)+\left(6-2\sqrt6\sqrt2+2\right)}=\blue 4}\)
\(\displaystyle{ a=\black 4\left( -\frac{\sqrt6+\sqrt2}{4}+\frac{\sqrt6-\sqrt2}{4}\cdot i\right)}\)
\(\displaystyle{ \cos15^o=\cos\frac{30^o}{2}=\sqrt{\frac{1+\cos30^o}{2}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2+\sqrt3}{4}}=\sqrt{\frac{8+4\sqrt3}{16}}=}\)
\(\displaystyle{ =\sqrt{\frac{6+2\sqrt2\sqrt2\sqrt3+2}{16}}=\sqrt{\frac{\left( \sqrt6\right)^2 +2\sqrt2\sqrt6+\left( \sqrt2\right)^2 }{16}}=\sqrt{\frac{\left( \sqrt6+\sqrt2\right)^2}{16}}}\)
\(\displaystyle{ \blue \cos15^o=\frac{\sqrt6+\sqrt2}{4}\ \ \ \ \ \ \ \sin15^o=\frac{\sqrt6-\sqrt2}{4}}\)
\(\displaystyle{ a=4\left( -\cos15^o+i\,\sin15^o\right)=-4\left( \cos15^o-i\,\sin15^o\right)=-4\sqrt[4]{\left( \cos15^o-i\,\sin15^o\right)^4}=}\)
\(\displaystyle{ =-4\sqrt[4]{\left[ \cos(4\cdot15^o)-i\,\sin(4\cdot15^o)\right]}=-4\sqrt[4]{\left( \cos60^o-i\,\sin60^o\right)}=-4\left( \cos60^o-i\,\sin60^o\right)^\frac14=}\)
\(\displaystyle{ =-2^\frac742^\frac14\left( \cos60^o-i\,\sin60^o\right)^\frac14=-2^\frac74\left[2\left( \cos60^o-i\,\sin60^o\right)\right]^\frac14=}\)
\(\displaystyle{ =-2\cdot2^\frac34\left[2\left( \frac12-i\frac{\sqrt3}{2}\right)\right]^\frac14=\ \red -2\cdot2^\frac34\left( 1-i\sqrt3\right)^\frac14}\)
\(\displaystyle{ |a|=\sqrt{\left( -\sqrt6-\sqrt2\right)^2+\left( \sqrt6-\sqrt2\right)^2}=\sqrt{\left(6+2\sqrt6\sqrt2+2\right)+\left(6-2\sqrt6\sqrt2+2\right)}=\blue 4}\)
\(\displaystyle{ a=\black 4\left( -\frac{\sqrt6+\sqrt2}{4}+\frac{\sqrt6-\sqrt2}{4}\cdot i\right)}\)
\(\displaystyle{ \cos15^o=\cos\frac{30^o}{2}=\sqrt{\frac{1+\cos30^o}{2}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2+\sqrt3}{4}}=\sqrt{\frac{8+4\sqrt3}{16}}=}\)
\(\displaystyle{ =\sqrt{\frac{6+2\sqrt2\sqrt2\sqrt3+2}{16}}=\sqrt{\frac{\left( \sqrt6\right)^2 +2\sqrt2\sqrt6+\left( \sqrt2\right)^2 }{16}}=\sqrt{\frac{\left( \sqrt6+\sqrt2\right)^2}{16}}}\)
\(\displaystyle{ \blue \cos15^o=\frac{\sqrt6+\sqrt2}{4}\ \ \ \ \ \ \ \sin15^o=\frac{\sqrt6-\sqrt2}{4}}\)
\(\displaystyle{ a=4\left( -\cos15^o+i\,\sin15^o\right)=-4\left( \cos15^o-i\,\sin15^o\right)=-4\sqrt[4]{\left( \cos15^o-i\,\sin15^o\right)^4}=}\)
\(\displaystyle{ =-4\sqrt[4]{\left[ \cos(4\cdot15^o)-i\,\sin(4\cdot15^o)\right]}=-4\sqrt[4]{\left( \cos60^o-i\,\sin60^o\right)}=-4\left( \cos60^o-i\,\sin60^o\right)^\frac14=}\)
\(\displaystyle{ =-2^\frac742^\frac14\left( \cos60^o-i\,\sin60^o\right)^\frac14=-2^\frac74\left[2\left( \cos60^o-i\,\sin60^o\right)\right]^\frac14=}\)
\(\displaystyle{ =-2\cdot2^\frac34\left[2\left( \frac12-i\frac{\sqrt3}{2}\right)\right]^\frac14=\ \red -2\cdot2^\frac34\left( 1-i\sqrt3\right)^\frac14}\)