高専の数学解説

第11章 11.12

(1)

z2=2iz^2 = 2i

z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) とおくと与式は

r2(cos⁡θ+isin⁡θ)2=2ir^2(\cos\theta + i\sin\theta)^2 = 2i r2(cos⁡2θ+isin⁡2θ)=2ir^2(\cos 2\theta + i\sin 2\theta) = 2i これから

r2=2r^2 = 2 r=2r = \sqrt{2}

2θ=π2+2kπ2\theta = \frac{\pi}{2} + 2k\pi θ=π4+kπ\theta = \frac{\pi}{4} + k\pi

k=0k = 0 のとき θ=π4\theta = \frac{\pi}{4} k=1k = 1 のとき θ=54π\theta = \frac{5}{4}\pi これから

z1=2(cos⁡π4+isin⁡π4)=1+iz_1 = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = 1 + i

z2=2(cos⁡54π+isin⁡54π)=−1−iz_2 = \sqrt{2}\left(\cos\frac{5}{4}\pi + i\sin\frac{5}{4}\pi\right) = -1 - i

(2)

z2=1+3iz^2 = 1 + \sqrt{3}i

z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) とおくと与式は

r2(cos⁡θ+isin⁡θ)2=1+3ir^2(\cos\theta + i\sin\theta)^2 = 1 + \sqrt{3}i r2(cos⁡2θ+isin⁡2θ)=2(cos⁡π3+isin⁡π3)r^2(\cos 2\theta + i\sin 2\theta) = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) これから

r=2r = \sqrt{2}

2θ=π3+2kπ2\theta = \frac{\pi}{3} + 2k\pi θ=π6+kπ\theta = \frac{\pi}{6} + k\pi

k=0k = 0 のとき θ=π6\theta = \frac{\pi}{6} k=1k = 1 のとき θ=76π\theta = \frac{7}{6}\pi これから

z1=2(cos⁡π6+isin⁡π6)=62+22iz_1 = \sqrt{2}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i

z2=2(cos⁡76π+isin⁡76π)=−62−22iz_2 = \sqrt{2}\left(\cos\frac{7}{6}\pi + i\sin\frac{7}{6}\pi\right) = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i

(3)

z3=−iz^3 = -i

z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta) と置くと与式は

r3(cos⁡3θ+isin⁡3θ)=−i=cos⁡32π+isin⁡32πr^3(\cos 3\theta + i \sin 3\theta) = -i = \cos \frac{3}{2}\pi + i \sin \frac{3}{2}\pi これから

r3=1r^3 = 1 r=1r = 1

3θ=32π+2kπ3\theta = \frac{3}{2}\pi + 2k\pi θ=12π+23kπ\theta = \frac{1}{2}\pi + \frac{2}{3}k\pi

k=0k = 0 のとき θ=π2\theta = \frac{\pi}{2} k=1k = 1 のとき θ=76π\theta = \frac{7}{6}\pi k=2k = 2 のとき θ=116π\theta = \frac{11}{6}\pi これから

z1=cos⁡12π+isin⁡12π=iz_1 = \cos \frac{1}{2}\pi + i \sin \frac{1}{2}\pi = i

z2=cos⁡76π+isin⁡76π=−32−12iz_2 = \cos \frac{7}{6}\pi + i \sin \frac{7}{6}\pi = -\frac{\sqrt{3}}{2} - \frac{1}{2}i

z3=cos⁡116π+isin⁡116π=32−12iz_3 = \cos \frac{11}{6}\pi + i \sin \frac{11}{6}\pi = \frac{\sqrt{3}}{2} - \frac{1}{2}i

(4)

z4=−8+83iz^4 = -8 + 8\sqrt{3}i

z=r(cos⁡θ+isin⁡θ)4=−8+83iz = r(\cos \theta + i \sin \theta)^4 = -8 + 8\sqrt{3}i r4(cos⁡4θ+isin⁡4θ)=8{2(cos⁡23π+isin⁡23π)}r^4(\cos 4\theta + i \sin 4\theta) = 8\{2(\cos \frac{2}{3}\pi + i \sin \frac{2}{3}\pi)\} これから

r4=16r^4 = 16 r=2r = 2

4θ=23π+2kπ4\theta = \frac{2}{3}\pi + 2k\pi θ=π6+12kπ\theta = \frac{\pi}{6} + \frac{1}{2}k\pi

k=0k = 0 のとき θ=π6\theta = \frac{\pi}{6} k=1k = 1 のとき θ=23π\theta = \frac{2}{3}\pi

k=2k = 2 のとき θ=76π\theta = \frac{7}{6}\pi k=3k = 3 のとき θ=53π\theta = \frac{5}{3}\pi これから

z1=2(cos⁡π6+isin⁡π6)=3+iz_1 = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = \sqrt{3} + i z2=2(cos⁡23π+isin⁡23π)=−1+3iz_2 = 2(\cos \frac{2}{3}\pi + i \sin \frac{2}{3}\pi) = -1 + \sqrt{3}i

z3=2(cos⁡76π+isin⁡76π)=−3−iz_3 = 2(\cos \frac{7}{6}\pi + i \sin \frac{7}{6}\pi) = -\sqrt{3} - i z4=2(cos⁡53π+isin⁡53π)=1−3iz_4 = 2(\cos \frac{5}{3}\pi + i \sin \frac{5}{3}\pi) = 1 - \sqrt{3}i

解説: r31bn1z