Let Z1 and Z2 be two complex numbers such that Z¯1=iZ¯2 and argz1z¯2=π. Then

Z¯1=iZ¯2 z1=-iz2 given that argZ1Z2¯=π arg-iZ2Z¯2=π , let argz2=θ -π2+θ+θ=π 2θ=3π2 argz2=θ=3π4⇒argz1=π4

Let Z1 and Z2 be two complex numbers such that Z¯1=iZ¯2 and argz1z¯2=π. Then

Z¯1=iZ¯2 z1=-iz2 given that argZ1Z2¯=π arg-iZ2Z¯2=π , let argz2=θ -π2+θ+θ=π 2θ=3π2 argz2=θ=3π4⇒argz1=π4

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Let $Z_{1}$ and $Z_{2}$ be two complex numbers such that ${\bar{Z}}_{1} = i {\bar{Z}}_{2}$ and arg $(\frac{z_{1}}{{\bar{z}}_{2}})$ $= π$ . Then

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arg $Z_{2} = \frac{π}{4}$

arg $z_{2} = - \frac{3 π}{4}$

arg $z_{1} = \frac{π}{4}$

arg $z_{1} = - \frac{3 π}{4}$

answer is C.

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Detailed Solution

${\bar{Z}}_{1} = {\bar{i Z}}_{2} z_{1} = - i z_{2} g i v e n t h a t a r g (\frac{Z_{1}}{\bar{Z_{2}}}) = π a r g (\frac{- i Z_{2}}{{\bar{Z}}_{2}}) = π, l e t \arg (z_{2}) = θ - \frac{π}{2} + θ + θ = π 2 θ = \frac{3 π}{2} \arg (z_{2}) = θ = \frac{3 π}{4} \Rightarrow \arg z_{1} = \frac{π}{4}$