What is (x+iy)^2+i(x+iy)^2 ?

The solution to (x+iy)^2+i(x+iy)^2 is (x^2-y^2-2xy)+(2xy+x^2-y^2)i

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simplify (x+iy)^2+i(x+iy)^2

simplify

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

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

Solution steps

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

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

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

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

Expand

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

= x^{2} + 2 x y i - y^{2} + x^{2} i - 2 x y - i y^{2}

Rewrite in standard complex form:

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

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

(y - 1) d x

V (3, - 2) y F (3, 0)

\frac{( 1 + cos ( θ ) ) ^{2}}{2}

x \cdot sin (y) \cdot d x + (x^{2} + 1) cos (y) \cdot d y

\frac{( x - 6 ) ( x + 1 )}{( x - 3 )}

What is (x+iy)^2+i(x+iy)^2 ?
The solution to (x+iy)^2+i(x+iy)^2 is (x^2-y^2-2xy)+(2xy+x^2-y^2)i