The problem is about complex numbers. a. Given $z_1 = \sqrt{2} + i\sqrt{2}$, find $\overline{z_1}$ (the complex conjugate of $z_1$). b. Find the modulus and argument of the complex number $z_1$. Write $z_1$ in trigonometric form. c. Show that $\overline{z_1}$ is a root of the equation $z^2 = 2(z\sqrt{2} - 2)$.

1. Problem Description

The problem is about complex numbers.

a. Given $z_1 = \sqrt{2} + i\sqrt{2}$, find $\overline{z_1}$ (the complex conjugate of $z_1$).

b. Find the modulus and argument of the complex number $z_1$. Write $z_1$ in trigonometric form.

c. Show that $\overline{z_1}$ is a root of the equation $z^2 = 2(z\sqrt{2} - 2)$.

2. Solution Steps

a. To find the complex conjugate of $z_1 = \sqrt{2} + i\sqrt{2}$, we simply change the sign of the imaginary part.

$\overline{z_1} = \sqrt{2} - i\sqrt{2}$

b. To find the modulus of $z_1$, we use the formula $|z| = \sqrt{a^2 + b^2}$, where $z = a + bi$.

$|z_1| = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.

To find the argument of $z_1$, we use the formula $\theta = \arctan(\frac{b}{a})$.

$\theta = \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \arctan(1) = \frac{\pi}{4}$.

So, $z_1 = 2(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$.

c. We want to show that $\overline{z_1} = \sqrt{2} - i\sqrt{2}$ is a root of the equation $z^2 = 2(z\sqrt{2} - 2)$.

We substitute $\overline{z_1}$ into the equation:

$(\sqrt{2} - i\sqrt{2})^2 = 2((\sqrt{2} - i\sqrt{2})\sqrt{2} - 2)$

$(\sqrt{2})^2 - 2(\sqrt{2})(i\sqrt{2}) + (i\sqrt{2})^2 = 2(2 - 2i - 2)$

$2 - 4i - 2 = 2(-2i)$

$-4i = -4i$.

Therefore, $\overline{z_1}$ is indeed a root of the given equation.

3. Final Answer

a. $\overline{z_1} = \sqrt{2} - i\sqrt{2}$

b. $|z_1| = 2$, $\arg(z_1) = \frac{\pi}{4}$, $z_1 = 2(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$

c. $\overline{z_1}$ is a root of the equation $z^2 = 2(z\sqrt{2} - 2)$.