If z1 = 1, z2 = i and A = Arg(z1 z2) B = Argz1z2, C = Arg(z1 + z2), D = Arg(z1 - z2) arrange A,B,C,D in ascending order

z1 = 1, z2 = i A=Arg z1 z2= Arg i=π2Hence, A=argz1·z2=argi=π2B=Argz1z2=Arg1i=Arg 1-Arg i=0-π2=-π2C=Argz1+z2=Arg1+iTan-1 1=π4D=Argz1-z2=Arg1-i=Tan-1(-1)-π4∴B<D<C<A

If z1 = 1, z2 = i and A = Arg(z1 z2) B = Argz1z2, C = Arg(z1 + z2), D = Arg(z1 - z2) arrange A,B,C,D in ascending order

z1 = 1, z2 = i A=Arg z1 z2= Arg i=π2Hence, A=argz1·z2=argi=π2B=Argz1z2=Arg1i=Arg 1-Arg i=0-π2=-π2C=Argz1+z2=Arg1+iTan-1 1=π4D=Argz1-z2=Arg1-i=Tan-1(-1)-π4∴B<D<C<A

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If $z_{1} = 1, z_{2} = i and A = Arg (z_{1} z_{2}) B = Arg (\frac{z_{1}}{z_{2}})$ , $C = Arg (z_{1} + z_{2}), D = Arg (z_{1} - z_{2})$ arrange A,B,C,D in ascending order

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A, C, D, B

B, D, C, A

B, C, D, A

B, A, D, C

answer is B.

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Detailed Solution

$z_{1} = 1, z_{2} = i$

$A = Arg (z_{1} z_{2}) = Arg i = \frac{π}{2}$

Hence, $A = a r g (z_{1} \cdot z_{2}) = a r g (i) = \frac{π}{2}$

$\begin{array}{rcl} B & = & Arg (\frac{z_{1}}{z_{2}}) = Arg (\frac{1}{i}) = Arg 1 - Arg i = 0 - \frac{π}{2} = - \frac{π}{2} \end{array}$

$\begin{array}{rcl} C & = & Arg (z_{1} + z_{2}) = Arg (1 + i) {Tan}^{- 1} 1 = \frac{π}{4} \\ D & = & Arg (z_{1} - z_{2}) = Arg (1 - i) = {Tan}^{- 1} (- 1) - \frac{π}{4} \end{array}$