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Find real x and y such that:
(x+yi)^2=x-yi

NEED HELP NOW PLEASE

User ManishChristian
by
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2 Answers

13 votes
13 votes

Answer:

y = sqrt(-2 x - 1/4) + 1/2 (2 i x + i) or y = 1/2 (2 i x + i) - sqrt((-2 x - 1/4))

x = sqrt((-2 i) y + 1/4) + 1/2 (-2 i y + 1) or x = 1/2 (-2 i y + 1) - sqrt(((-2 i) y + 1/4))

Explanation:

Solve for y:

(x + (0 + i) y)^2 = x + (0 - i) y

Subtract -i y + x from both sides:

-x + (i y + x)^2 + i y = 0

Expand and collect in terms of y:

-x + x^2 + y (2 i x + i) - y^2 = 0

Multiply both sides by -1:

x - x^2 + y (-2 i x - i) + y^2 = 0

Subtract -x^2 + x from both sides:

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

Add 1/4 (-2 i x - i)^2 to both sides:

1/4 (-2 i x - i)^2 + y (-2 i x - i) + y^2 = 1/4 (-2 i x - i)^2 - x + x^2

Write the left hand side as a square:

(1/2 (-2 i x - i) + y)^2 = 1/4 (-2 i x - i)^2 - x + x^2

Take the square root of both sides:

1/2 (-2 i x - i) + y = sqrt(1/4 (-2 i x - i)^2 - x + x^2) or 1/2 (-2 i x - i) + y = -sqrt(1/4 (-2 i x - i)^2 - x + x^2)

Subtract 1/2 (-2 i x - i) from both sides:

y = 1/2 (2 i x + i) + sqrt((((-2 i) x - i)^2)/4 - x + x^2) or 1/2 (-2 i x - i) + y = -sqrt(1/4 (-2 i x - i)^2 - x + x^2)

1/4 (-2 i x - i)^2 - x + x^2 = -2 x - 1/4:

y = 1/2 (2 i x + i) + sqrt((-2 x - 1/4)) or 1/2 (-2 i x - i) + y = -sqrt(1/4 (-2 i x - i)^2 - x + x^2)

Subtract 1/2 (-2 i x - i) from both sides:

y = sqrt(-2 x - 1/4) + 1/2 (2 i x + i) or y = 1/2 (2 i x + i) - sqrt((((-2 i) x - i)^2)/4 - x + x^2)

1/4 (-2 i x - i)^2 - x + x^2 = -2 x - 1/4:

Answer: y = sqrt(-2 x - 1/4) + 1/2 (2 i x + i) or y = 1/2 (2 i x + i) - sqrt((-2 x - 1/4))

________________________

Solve for x:

(x + (0 + i) y)^2 = x + (0 - i) y

Subtract -i y + x from both sides:

-x + (i y + x)^2 + i y = 0

Expand and collect in terms of x:

x^2 + x (2 i y - 1) + i y - y^2 = 0

Subtract -y^2 + i y from both sides:

x^2 + x (2 i y - 1) = y^2 - i y

Add 1/4 (2 i y - 1)^2 to both sides:

x^2 + x (2 i y - 1) + 1/4 (2 i y - 1)^2 = 1/4 (2 i y - 1)^2 - i y + y^2

Write the left hand side as a square:

(x + 1/2 (2 i y - 1))^2 = 1/4 (2 i y - 1)^2 - i y + y^2

Take the square root of both sides:

x + 1/2 (2 i y - 1) = sqrt(1/4 (2 i y - 1)^2 - i y + y^2) or x + 1/2 (2 i y - 1) = -sqrt(1/4 (2 i y - 1)^2 - i y + y^2)

Subtract 1/2 (2 i y - 1) from both sides:

x = 1/2 (-2 i y + 1) + sqrt((((2 i) y - 1)^2)/4 - i y + y^2) or x + 1/2 (2 i y - 1) = -sqrt(1/4 (2 i y - 1)^2 - i y + y^2)

1/4 (2 i y - 1)^2 - i y + y^2 = -2 i y + 1/4:

x = 1/2 (-2 i y + 1) + sqrt(((-2 i) y + 1/4)) or x + 1/2 (2 i y - 1) = -sqrt(1/4 (2 i y - 1)^2 - i y + y^2)

Subtract 1/2 (2 i y - 1) from both sides:

x = sqrt((-2 i) y + 1/4) + 1/2 (-2 i y + 1) or x = 1/2 (-2 i y + 1) - sqrt(((2 i y - 1)^2)/4 - i y + y^2)

1/4 (2 i y - 1)^2 - i y + y^2 = -2 i y + 1/4:

Answer: x = sqrt((-2 i) y + 1/4) + 1/2 (-2 i y + 1) or x = 1/2 (-2 i y + 1) - sqrt(((-2 i) y + 1/4))

User Theplau
by
3.0k points
24 votes
24 votes

Answer:

Explanation:

Find real x and y such that: (x+yi)^2=x-yi NEED HELP NOW PLEASE-example-1
User Fravolt
by
3.3k points
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