(B) Given $z = x+iy$,we have $z-2-3i = (x-2) + i(y-3)$.Since the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,we have $\arg((x-2) + i(y-3)) = \frac{\pi}{4

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(B) Given $z = x+iy$,we have $z-2-3i = (x-2) + i(y-3)$.Since the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,we have $\arg((x-2) + i(y-3)) = \frac{\pi}{4}$.This implies $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.Taking the tangent on both sides,we get $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.Therefore,$y-3 = x-2$,which simplifies to $x-y+1=0$.

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

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If the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,then the locus of $z=x+iy$ is:

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A
$x-y-1=0$
B
$x-y+1=0$
C
$x+y-1=0$
D
$x+y+1=0$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Geometry of complex numbers

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