What is the answer to a=(ax-Bx)(ai-Bi)+(-ay)(-aj) ?

The answer to a=(ax-Bx)(ai-Bi)+(-ay)(-aj) is x=0,y= 1/(aj),\quad a\ne 0

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a=(ax-Bx)(ai-Bi)+(-ay)(-aj)

a = (a x - B x) (ai - B i) + (- a y) (- aj)

x = 0, y = \frac{1}{aj}, a \neq = 0

Solution steps

a = (a x - B x) (ai - B i) + (- a y) (- aj)

Rewrite

a

in standard complex form:

a + 0 i

a + 0 i = (a x - B x) (ai - B i) + (- a y) (- aj)

Rewrite

(a x - B x) (ai - B i) + (- a y) (- aj)

in standard complex form:

a^{2} j y + (a^{2} x - 2 a B x + B^{2} x) i

a + 0 i = a^{2} j y + (a^{2} x - 2 a B x + B^{2} x) i

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

[a = a^{2} j y 0 = a^{2} x - 2 a B x + B^{2} x]

[a = a^{2} j y 0 = a^{2} x - 2 a B x + B^{2} x] : x = 0, y = \frac{1}{aj}, a \neq = 0

x = 0, y = \frac{1}{aj}, a \neq = 0

w^{2} + 4 w + (4 + 2 i) = 0

\frac{3 z + i}{( 2 + i ) z + 1} = i

5 x^{2} - 11 i x - 2 = 0

\frac{2 ( x + i y )}{3 - ( x - i y )} = i

(3 x - y) + (x + y) i = 12

What is the answer to a=(ax-Bx)(ai-Bi)+(-ay)(-aj) ?
The answer to a=(ax-Bx)(ai-Bi)+(-ay)(-aj) is x=0,y= 1/(aj),\quad a\ne 0