(D) Given $\left|\frac{z}{1+i}\right|=2$. Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$. Substituting $z=x+iy$,we get $

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(D) Given $\left|\frac{z}{1+i}\right|=2$. Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$. Substituting $z=x+iy$,we get $|x+iy| = 2\sqrt{2}$. Squaring both sides,we get $x^2+y^2 = (2\sqrt{2})^2 = 8$. This represents a circle $x^2+y^2 = 8$ with centre $C \equiv (0,0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

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If $\left|\frac{z}{1+i}\right|=2$,where $z=x+iy$ and $i=\sqrt{-1}$ represents a circle,then the centre $C$ and radius $r$ of the circle are:

MHT CET 2024 All MHT CET

A
$C \equiv(3,0), r=4$
B
$C \equiv(6,0), r=2$
C
$C \equiv(0,3), r=8$
D
$C \equiv(0,0), r=2\sqrt{2}$

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4-1.Complex numbers

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If $\left|\frac{z}{1+i}\right|=2$,where $z=x+iy$ and $i=\sqrt{-1}$ represents a circle,then the centre $C$ and radius $r$ of the circle are:

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