Let \(\mathrm{z}=1+\mathrm{i}\) and \(\mathrm{z}_{1}=\frac{1+\mathrm{i} \overline{\mathrm{z}}}{\overline{\mathrm{z}}(1-\mathrm{z})+\frac{1}{\mathrm{z}}}\). Then \(\frac{12}{\pi} \arg \left(\mathrm{z}_{1}\right)\) is equal to ______.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Complex Numbers

Let

z = 1 + i

and

z_{1} = \frac{1 + i z}{z ( 1 - z ) + \frac{1}{z}}

. Then

\frac{12}{π} ar g (z_{1})

is equal to ______.

Answer: 9

With

z = 1 + i

and

\overset{z}{ˉ} = 1 - i

, the numerator is

1 + i \overset{z}{ˉ} = 1 + i (1 - i) = 2 + i

. For the denominator,

\overset{z}{ˉ} (1 - z) = (1 - i) (- i) = - 1 - i

and

\frac{1}{z} = \frac{1 - i}{2}

, so the denominator is

- 1 - i + \frac{1 - i}{2} = \frac{- 1 - 3 i}{2}

. Hence

z_{1} = \frac{2 + i}{\frac{- 1 - 3 i}{2}} = \frac{2 ( 2 + i )}{- 1 - 3 i} = \frac{( 4 + 2 i ) ( - 1 + 3 i )}{( - 1 ) ^{2} + 3 ^{2}} = \frac{- 10 + 10 i}{10} = - 1 + i .

This lies in the second quadrant with argument

\frac{3 π}{4}

, so

\frac{12}{π} ar g (z_{1}) = \frac{12}{π} \cdot \frac{3 π}{4} = 9.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer