If z=1+i3 then match the following.IArg(z) a5π6IIArg(iz) b-π3III Arg(z¯)cπIV Arg(iz¯) dπ6 eπ3

z=1+i3⇒argz=Tan-131=π3iz=-3+i⇒Argiz=π-Tan-113=π-π6=5π6Argz=Arg1-i3=-Tan-13=-π3iz=i1-i3=3+i ⇒Argiz=Tan-113=π6

If z=1+i3 then match the following.IArg(z) a5π6IIArg(iz) b-π3III Arg(z¯)cπIV Arg(iz¯) dπ6 eπ3

z=1+i3⇒argz=Tan-131=π3iz=-3+i⇒Argiz=π-Tan-113=π-π6=5π6Argz=Arg1-i3=-Tan-13=-π3iz=i1-i3=3+i ⇒Argiz=Tan-113=π6

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If $z = 1 + i \sqrt{3}$ then match the following.
I $Arg (z)$ a $\frac{5 π}{6}$
II $Arg (iz)$ b $\frac{- π}{3}$
III $Arg (\bar{z})$ c $π$
IV $Arg (i \bar{z})$ d $\frac{π}{6}$
e $\frac{π}{3}$

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e, a, b, c

e, b, a, c

e, b, a, d

e, a, b, d

answer is D.

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Detailed Solution

$z = 1 + i \sqrt{3} \Rightarrow a r g (z) = T a n^{- 1} \frac{\sqrt{3}}{1} = \frac{π}{3}$

$i z = - \sqrt{3} + i \Rightarrow A r g (i z) = π - {Tan}^{- 1} \frac{1}{\sqrt{3}} = π - \frac{π}{6} = \frac{5 π}{6}$

$A r g (z) = A r g (1 - i \sqrt{3}) = - T a n^{- 1} (\sqrt{3}) = - \frac{π}{3}$

$i z = i (1 - i \sqrt{3}) = \sqrt{3} + i \Rightarrow A r g (i z) = T a n^{- 1} \frac{1}{\sqrt{3}} = \frac{π}{6}$

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I	$Arg (z)$	a	$\frac{5 π}{6}$
II	$Arg (iz)$	b	$\frac{- π}{3}$
III	$Arg (\bar{z})$	c	$π$
IV	$Arg (i \bar{z})$	d	$\frac{π}{6}$
		e	$\frac{π}{3}$

If z=1+i3 then match the following.IArg(z) a5π6IIArg(iz) b-π3III Arg(z¯)cπIV Arg(iz¯) dπ6 eπ3

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If $z = 1 + i \sqrt{3}$ then match the following.
I $Arg (z)$ a $\frac{5 π}{6}$
II $Arg (iz)$ b $\frac{- π}{3}$
III $Arg (\bar{z})$ c $π$
IV $Arg (i \bar{z})$ d $\frac{π}{6}$
e $\frac{π}{3}$