(C) Given $|1 + iz| = |1 - iz|$.Substitute $z = x + iy$:$|1 + i(x + iy)| = |1 - i(x + iy)|$$|1 + ix - y| = |1 - ix + y|$$|(1 - y) + ix| = |(1 + y) - i

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(C) Given $|1 + iz| = |1 - iz|$.Substitute $z = x + iy$:$|1 + i(x + iy)| = |1 - i(x + iy)|$$|1 + ix - y| = |1 - ix + y|$$|(1 - y) + ix| = |(1 + y) - ix|$Squaring both sides:$(1 - y)^2 + x^2 = (1 + y)^2 + x^2$$1 - 2y + y^2 + x^2 = 1 + 2y + y^2 + x^2$$-2y = 2y$$4y = 0 \Rightarrow y = 0$.Since $z = x + iy$ and $y = 0$, we have $z = x$.Also, $\bar{z} = x - iy = x - i(0) = x$.Therefore, $z = \bar{z}$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Geometry of complex numbers

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If $z = x + iy$ is a complex number and $|1 + iz| = |1 - iz|$, then

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A
$\operatorname{Re}(z) > 0$
B
$|z| = 1$
C
$z = \bar{z}$
D
$z = -\bar{z}$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Geometry of complex numbers

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