zz=13-13i

解

zz = 13 - 13 i

z = \frac{13 ( 1 + 2 )}{2} - \frac{13}{2 1 + 2} i, z = - \frac{13 ( 1 + 2 )}{2} + \frac{13}{2 1 + 2} i

解答ステップ

zz = 13 - 13 i

代用

z = x + y i

(x + y i) (x + y i) = 13 - 13 i

簡素化

(x + y i) (x + y i) : (x^{2} - y^{2}) + 2 i x y

(x^{2} - y^{2}) + 2 i x y = 13 - 13 i

複素数は, その実数部と虚数部が等しい場合にのみ等しくなるequation系として書き換える:

[x^{2} - y^{2} = 13 2 x y = - 13]

[x^{2} - y^{2} = 13 2 x y = - 13] : x = \frac{13 ( 1 + 2 )}{2}, x = - \frac{13 ( 1 + 2 )}{2}, y = - \frac{13}{2 1 + 2} y = \frac{13}{2 1 + 2}

代用を戻す

z = x + y i

z = \frac{13 ( 1 + 2 )}{2} - \frac{13}{2 1 + 2} i, z = - \frac{13 ( 1 + 2 )}{2} + \frac{13}{2 1 + 2} i