The problem is to solve the quadratic equation $ix^2 + 2(1+i)x + 1 = 0$ for $x$.

1. Problem Description

The problem is to solve the quadratic equation $ix^2 + 2(1+i)x + 1 = 0$ for $x$.

2. Solution Steps

We can use the quadratic formula to solve for $x$. The quadratic formula for $ax^2 + bx + c = 0$ is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = i$, $b = 2(1+i)$ and $c = 1$.

Substituting these values into the quadratic formula:

$x = \frac{-2(1+i) \pm \sqrt{(2(1+i))^2 - 4(i)(1)}}{2i}$

First, let's simplify the expression under the square root:

$(2(1+i))^2 - 4(i)(1) = 4(1+i)^2 - 4i = 4(1 + 2i + i^2) - 4i = 4(1 + 2i - 1) - 4i = 4(2i) - 4i = 8i - 4i = 4i$

So,

$x = \frac{-2(1+i) \pm \sqrt{4i}}{2i}$

We need to find the square root of $4i$. Let $\sqrt{4i} = a + bi$. Then $(a+bi)^2 = 4i$, which means $a^2 + 2abi - b^2 = 4i$.

Equating the real and imaginary parts, we have:

$a^2 - b^2 = 0$ and $2ab = 4$, which implies $ab=2$.

From the first equation, $a^2 = b^2$, so $a = b$ or $a = -b$. Since $ab = 2 > 0$, $a$ and $b$ must have the same sign, so we have $a = b$.

Substituting $a = b$ into $ab = 2$, we get $a^2 = 2$, so $a = \sqrt{2}$ or $a = -\sqrt{2}$.

Then $b = \sqrt{2}$ or $b = -\sqrt{2}$.

Therefore, $\sqrt{4i} = \sqrt{2} + i\sqrt{2}$ or $\sqrt{4i} = -\sqrt{2} - i\sqrt{2}$. We can take $\sqrt{4i} = \sqrt{2} + i\sqrt{2} = \sqrt{2}(1+i)$.

Substituting this back into the equation for $x$:

$x = \frac{-2(1+i) \pm \sqrt{2}(1+i)}{2i} = \frac{(1+i)(-2 \pm \sqrt{2})}{2i}$

$x = \frac{(1+i)(-2 \pm \sqrt{2})}{2i} \cdot \frac{-i}{-i} = \frac{(1+i)(-2 \pm \sqrt{2})(-i)}{2} = \frac{(1+i)(2i \mp \sqrt{2}i)}{2} = \frac{2i \mp \sqrt{2}i - 2 \mp \sqrt{2}}{2}$

$x = \frac{-2 \mp \sqrt{2} + i(2 \mp \sqrt{2})}{2} = -1 \mp \frac{\sqrt{2}}{2} + i (1 \mp \frac{\sqrt{2}}{2})$

So the two possible values of $x$ are $x = -1 - \frac{\sqrt{2}}{2} + i(1 - \frac{\sqrt{2}}{2})$ and $x = -1 + \frac{\sqrt{2}}{2} + i(1 + \frac{\sqrt{2}}{2})$.

$x = \frac{-2(1+i) \pm \sqrt{4i}}{2i} = \frac{-2(1+i) \pm 2\sqrt{i}}{2i} = \frac{-(1+i) \pm \sqrt{i}}{i}$

Let $\sqrt{i} = \frac{1+i}{\sqrt{2}}$, then

$x = \frac{-1-i \pm \frac{1+i}{\sqrt{2}}}{i} = (-1-i \pm \frac{1+i}{\sqrt{2}}) \frac{-i}{-i^2} = (-1-i \pm \frac{1+i}{\sqrt{2}})(-i) = i + i^2 \mp \frac{i-1}{\sqrt{2}} = i - 1 \mp \frac{i-1}{\sqrt{2}} = -1 \mp \frac{-1}{\sqrt{2}} + i \mp \frac{1}{\sqrt{2}} = -1 \pm \frac{1}{\sqrt{2}} + i(1 \pm \frac{1}{\sqrt{2}})$

$x = -1 \pm \frac{\sqrt{2}}{2} + i(1 \pm \frac{\sqrt{2}}{2})$

3. Final Answer

The solutions are $x = -1 + \frac{\sqrt{2}}{2} + i(1 + \frac{\sqrt{2}}{2})$ and $x = -1 - \frac{\sqrt{2}}{2} + i(1 - \frac{\sqrt{2}}{2})$.