The complex number satisfying $\arg\left(z+i\right)=\frac{\pi}{4}$ and $\arg\left(2z+3-2i\right)=\frac{3\pi}{4}$ simultaneously, is :
$\frac{1}{4}+\frac{3}{4}i$
$\frac{1}{4}+\frac{3}{4}i$
$-\frac{1}{4}+\frac{3}{4}i$
None of these
Step-by-Step Solution
Key Concept: The argument conditions impose linear constraints on the real and imaginary parts of $z$ that must be solved simultaneously while checking the quadrant requirements.
Let $z = x + yi$. From $\arg(z+i) = \frac{\pi}{4}$, we have $z + i = x + (y+1)i$, so $\frac{y+1}{x} = 1$, giving $y + 1 = x$. From $\arg(2z + 3 - 2i) = \frac{3\pi}{4}$, we have $2z + 3 - 2i = (2x+3) + (2y-2)i$, so $\frac{2y-2}{2x+3} = -1$, giving $2y - 2 = -(2x+3)$, or $2y = -2x - 1$. Solving simultaneously: $y = x - 1$ and $2(x-1) = -2x - 1$ yields $2x - 2 = -2x - 1$, so $4x = 1$ and $x = \frac{1}{4}$, thus $y = -\frac{3}{4}$. However, checking the first condition requires $x > 0$ and $y + 1 > 0$, and the second requires $2x + 3 > 0$ and $2y - 2 < 0$, which gives $z = \frac{1}{4} - \frac{3}{4}i$. Since this doesn't match any of the given options, the answer is None of these.
Correct Answer: 4