複素数の割り算 $\frac{-2+2i}{-1+\sqrt{3}i}$ を、オイラーの公式を用いて計算します。代数学複素数オイラーの公式極形式三角関数2025/7/91. 問題の内容複素数の割り算 −2+2i−1+3i\frac{-2+2i}{-1+\sqrt{3}i}−1+3i−2+2i を、オイラーの公式を用いて計算します。2. 解き方の手順まず、分子と分母をそれぞれ極形式で表します。分子: −2+2i-2 + 2i−2+2i絶対値 r1=(−2)2+22=4+4=8=22r_1 = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}r1=(−2)2+22=4+4=8=22偏角 θ1=arctan(2−2)+π=arctan(−1)+π=−π4+π=3π4\theta_1 = \arctan\left(\frac{2}{-2}\right) + \pi = \arctan(-1) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}θ1=arctan(−22)+π=arctan(−1)+π=−4π+π=43πよって、 −2+2i=22ei3π4-2+2i = 2\sqrt{2}e^{i\frac{3\pi}{4}}−2+2i=22ei43π分母: −1+3i-1 + \sqrt{3}i−1+3i絶対値 r2=(−1)2+(3)2=1+3=4=2r_2 = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2r2=(−1)2+(3)2=1+3=4=2偏角 θ2=arctan(3−1)+π=arctan(−3)+π=−π3+π=2π3\theta_2 = \arctan\left(\frac{\sqrt{3}}{-1}\right) + \pi = \arctan(-\sqrt{3}) + \pi = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}θ2=arctan(−13)+π=arctan(−3)+π=−3π+π=32πよって、 −1+3i=2ei2π3-1 + \sqrt{3}i = 2e^{i\frac{2\pi}{3}}−1+3i=2ei32πしたがって、−2+2i−1+3i=22ei3π42ei2π3=2ei(3π4−2π3)=2ei(9π−8π12)=2eiπ12\frac{-2+2i}{-1+\sqrt{3}i} = \frac{2\sqrt{2}e^{i\frac{3\pi}{4}}}{2e^{i\frac{2\pi}{3}}} = \sqrt{2}e^{i(\frac{3\pi}{4} - \frac{2\pi}{3})} = \sqrt{2}e^{i(\frac{9\pi - 8\pi}{12})} = \sqrt{2}e^{i\frac{\pi}{12}}−1+3i−2+2i=2ei32π22ei43π=2ei(43π−32π)=2ei(129π−8π)=2ei12π次に、指数形式から直交形式に戻します。eiπ12=cos(π12)+isin(π12)e^{i\frac{\pi}{12}} = \cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)ei12π=cos(12π)+isin(12π)cos(π12)=cos(π3−π4)=cos(π3)cos(π4)+sin(π3)sin(π4)=1222+3222=2+64\cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right) = \frac{1}{2}\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}cos(12π)=cos(3π−4π)=cos(3π)cos(4π)+sin(3π)sin(4π)=2122+2322=42+6sin(π12)=sin(π3−π4)=sin(π3)cos(π4)−cos(π3)sin(π4)=3222−1222=6−24\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} - \frac{1}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}sin(12π)=sin(3π−4π)=sin(3π)cos(4π)−cos(3π)sin(4π)=2322−2122=46−2よって、−2+2i−1+3i=2(2+64+i6−24)=2+234+i23−24=1+32+i3−12\frac{-2+2i}{-1+\sqrt{3}i} = \sqrt{2}\left(\frac{\sqrt{2} + \sqrt{6}}{4} + i\frac{\sqrt{6} - \sqrt{2}}{4}\right) = \frac{2 + 2\sqrt{3}}{4} + i\frac{2\sqrt{3} - 2}{4} = \frac{1+\sqrt{3}}{2} + i\frac{\sqrt{3}-1}{2}−1+3i−2+2i=2(42+6+i46−2)=42+23+i423−2=21+3+i23−13. 最終的な答え1+32+i3−12\frac{1+\sqrt{3}}{2} + i\frac{\sqrt{3}-1}{2}21+3+i23−1