For a complex number z, if `z^(2)+barz-z=4i` and z does not lie in the first quadrant, then `("where "i^(2)=-1)`

Options

A. `|z|=sqrt2`
B. `|z|=2sqrt2`
C. `arg(z)=(-pi)/(4)`
D. `arg(z)=(3pi)/(4)`

Answer

To solve the problem, we start with the given equation for the complex number \( z \):

\[ z^2 + \bar{z} - z = 4i \]

where \( \bar{z} \) is the conjugate of \( z \). We will express \( z \) in terms of its real and imaginary parts. Let:

\[ z = x + iy \]

Then, the conjugate \( \bar{z} \) is:

\[ \bar{z} = x - iy \]

Now we can substitute \( z \) and \( \bar{z} \) into the equation:

\[ (x + iy)^2 + (x - iy) - (x + iy) = 4i \]

Expanding \( (x + iy)^2 \):

\[ (x + iy)^2 = x^2 + 2xyi - y^2 \]

Now substituting this back into the equation gives:

\[ (x^2 - y^2 + 2xyi) + (x - iy) - (x + iy) = 4i \]

Simplifying the left-hand side:

\[ x^2 - y^2 + 2xyi + x - iy - x - iy = 4i \]

This simplifies to:

\[ x^2 - y^2 + (2xy - 2y)i = 4i \]

Now, we can separate the real and imaginary parts:

1. Real part: \( x^2 - y^2 = 0 \)
2. Imaginary part: \( 2xy - 2y = 4 \)

From the first equation \( x^2 - y^2 = 0 \), we can conclude:

\[ x^2 = y^2 \implies x = y \text{ or } x = -y \]

Now, let's consider the second equation:

\[ 2y(x - 1) = 4 \implies y(x - 1) = 2 \]

### Case 1: \( x = y \)

Substituting \( x = y \) into \( y(x - 1) = 2 \):

\[ y(y - 1) = 2 \]

This leads to:

\[ y^2 - y - 2 = 0 \]

Factoring gives:

\[ (y - 2)(y + 1) = 0 \]

Thus, \( y = 2 \) or \( y = -1 \). If \( y = 2 \), then \( x = 2 \) (which is in the first quadrant, not allowed). If \( y = -1 \), then \( x = -1 \).

### Case 2: \( x = -y \)

Substituting \( x = -y \) into \( y(x - 1) = 2 \):

\[ y(-y - 1) = 2 \]

This leads to:

\[ -y^2 - y - 2 = 0 \implies y^2 + y + 2 = 0 \]

This does not yield real solutions. Thus, we only have valid solutions from Case 1.

### Conclusion

We have \( x = -1 \) and \( y = -1 \), so:

\[ z = -1 - i \]

Now we calculate the modulus and argument of \( z \):

1. **Modulus**:

\[ |z| = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Argument**:

\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

Since \( z \) is in the third quadrant, the argument is:

\[ \theta = -\frac{\pi}{4} \]

### Final Answers

- Modulus of \( z \): \( \sqrt{2} \)
- Argument of \( z \): \( -\frac{\pi}{4} \)