Question
Find the image of the circle under the mapping
. What happens when
?
Answer :
Word Count : 528
Let ( z = x + iy ), where ( x, y \in \mathbb{R} ). The mapping is [ w = \frac{z - i}{z + i} = \frac{x + i(y-1)}{x + i(y+1)}. ] To find the image, write ( w = u + iv ), where ( u, v \in \mathbb{R} ). Multiply numerator and denominator by the conjugate of the denominator: [ w = \frac{x + i(y-1)}{x + i(y+1)} \cdot \frac{x - i(y+1)}{x - i(y+1)} = \frac{x^2 + (y-1)(y+1) + i[-x(y+1) + x(y-1)]}{x^2 + (y+1)^2}. ] Simplify numerator: * Real part: ( x^2 + y^2 - 1 ) * Imaginary part: ( i[-x(y+1) + ___ _____ ___ ____ ________.
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Let ( z = x + iy ), where ( x, y \in \mathbb{R} ). The mapping is [ w = \frac{z - i}{z + i} = \frac{x + i(y-1)}{x + i(y+1)}. ] To find the image, write ( w = u + iv ), where ( u, v \in \mathbb{R} ). Multiply numerator and denominator by the conjugate of the denominator: [ w = \frac{x + i(y-1)}{x + i(y+1)} \cdot \frac{x - i(y+1)}{x - i(y+1)} = \frac{x^2 + (y-1)(y+1) + i[-x(y+1) + x(y-1)]}{x^2 + (y+1)^2}. ] Simplify numerator: * Real part: ( x^2 + y^2 - 1 ) * Imaginary part: ( i[-x(y+1) + ___ _____ ___ ____ ________.
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