1.次の式を計算しなさい。
(1)\((-1+2i)+(3-4i)\)
\(=2-2i\)
(2)\((2-3i)-(4-2i)\)
\(=-2-i\)
(3)\((3+2i)(4-2i)\)
\(=12-6i+8i-4i^2\)
\(=16+2i\)
(4)\((2+3i)^2\)
\(=4+12i+9i^2\)
\(=-5+12i\)
(5)\((6-2i)(-3+4i)\)
\(=-18+24i+6i-8i^2\)
\(=-10+30i\)
(6)\(\displaystyle \frac{2}{1-i}\)
\(\displaystyle =\frac{2(1+i)}{(1-i)(1+i)}\)
\(\displaystyle =\frac{2+2i}{1-i^2}\)
\(\displaystyle =\frac{2+2i}{2}\)
\(=1+i\)
(7)\(\displaystyle \frac{-4+2i}{2i}\)
\(\displaystyle =\frac{i(-4+2i)}{2i^2}\)
\(\displaystyle =\frac{-4i+2i^2}{-2}\)
\(\displaystyle =\frac{-2-4i}{-2}\)
\(=1+2i\)
(8)\(\displaystyle \frac{5-3i}{1+5i}\)
\(\displaystyle =\frac{(5-3i)(1-5i)}{(1+5i)(1-5i)}\)
\(\displaystyle =\frac{5-25i-3i+15i^2}{1-25i^2}\)
\(\displaystyle =\frac{-10-28i}{26}\)
\(\displaystyle =\frac{-5-14i}{13}\)
(9)\(\displaystyle \frac{1+2i}{3+2i}+\frac{2+i}{3-2i}\)
\(\displaystyle =\frac{(1+2i)(3-2i)+(2+i)(3+2i)}{(3+2i)(3-2i)}\)
\(\displaystyle =\frac{3-2i+6i-4i^2+6+4i+3i+2i^2}{9-4i^2}\)
\(\displaystyle =\frac{11+11i}{13}\)
(10)\(\displaystyle \frac{2+5i}{3-2i}\)
\(\displaystyle =\frac{(2+5i)(3+2i)}{(3-2i)(3+2i)}\)
\(\displaystyle =\frac{6+4i+15i+10i^2}{9-4i^2}\)
\(\displaystyle =\frac{-4+19i}{13}\)
2.次の等式をみたす実数\(x,y\)を求めなさい。
(1)\((x-3)+(x+y)i=0\)
\(x-3,x+y\)は実数なので、
\(\left\{\begin{array}{l}x-3=0 \\ x+y=0\end{array}\right.\)
これを解くと、
\(x=3,y=-3\)
(2)\((x+yi)(2-i)=4-7i)\)
\(2x-xi+2yi-yi^2=4-7i\)
\((2x+y)+(-x+2y)i=4-7i\)
\(2x+y,-x+2y\)は実数なので、
\(\left\{\begin{array}{l}2x+y=4 \\ -x+2y=7\end{array}\right.\)
これを解くと、
\(x=3,y=-2\)
(3)\((4+3i)x+(1+2i)y+5=0\)
\((4x+y+5)+(3x+2y)i=0\)
\(4x+y+5,3x+2y\)は実数なので、
\(\left\{\begin{array}{l}4x+y+5=0 \\ 3x+2y=0\end{array}\right.\)
これを解くと、
\(x=-2,y=3\)
(4)\((5+2i)x+(2-2i)y=16-2i\)
\((5x+2y)+(2x-2y)i=16-2i\)
\(5x+2y,2x-2y\)は実数なので、
\(\left\{\begin{array}{l}5x+2y=16 \\ 2x-2y=-2\end{array}\right.\)
これを解くと、
\(x=2,y=3\)
(5)\((3+2i)(2x-yi)=4+7i\)
\((6x+2y)+(4x-3y)i=4+7i\)
\(6x+2y,4x-3y\)は実数なので、
\(\left\{\begin{array}{l}6x+2y=4 \\ 4x-3y=7\end{array}\right.\)
これを解くと、
\(x=1,y=-1\)
3.次の問いに答えなさい。
(1)\(-18\)の平方根を答えなさい。
\(\pm3\sqrt{2}i\)
(2)\(-8\)の平方根を答えなさい。
\(\pm2\sqrt{2}i\)
4.次の式を計算しなさい。
(1)\(\sqrt{-2}-\sqrt{-8}\)
\(=\sqrt{2}i-2\sqrt{2}i\)
\(=-\sqrt{2}i\)
(2)\(\sqrt{-12}\sqrt{-6}\)
\(=2\sqrt{3}i\sqrt{6}i\)
\(=2\sqrt{18}i^2\)
\(=-6\sqrt{2}\)
(3)\(\displaystyle \frac{\sqrt{24}}{\sqrt{-3}}\)
\(\displaystyle =\frac{2\sqrt{6}}{\sqrt{3}i}\)
\(\displaystyle =\frac{2\sqrt{2}i}{i^2}\)
\(=-2\sqrt{2}i\)
(4)\((\sqrt{3}+\sqrt{-2})(\sqrt{3}-\sqrt{-2})\)
\(=(\sqrt{3}+\sqrt{2}i)(\sqrt{3}-\sqrt{2}i)\)
\(=3-2i^2\)
\(=5\)
(5)\(\displaystyle \frac{2-\sqrt{-3}}{2+\sqrt{-3}}\)
\(\displaystyle =\frac{2-\sqrt{3}i}{2+\sqrt{3}i}\)
\(\displaystyle =\frac{(2-\sqrt{3}i)^2}{(2+\sqrt{3}i)(2-\sqrt{3}i)}\)
\(\displaystyle =\frac{4-4\sqrt{3}i+3i^2}{(4-3i^2}\)
\(\displaystyle =\frac{1-4\sqrt{3}i}{7}\)