What is the answer to zz=13-13i ?

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zz=13-13i

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

Solution steps

zz = 13 - 13 i

Substitute

z = x + y i

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

Simplify

(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

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[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}

Substitute back

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

x^{2} + 1 = i

a^{2} + b^{2} + a - bi = 11 + 3 i

t, a = \frac{V f - Vi}{t}

\frac{1}{z} = 1 - i

(x + y) + i (2 x - 3 y) = 2 + i