To solve the given problem, let us analyze and prove the equation:

Given:

• A complex number $z = x + iy$, where $x$ and $y$ are real.
• $z^2 = a + ib$, where $a$ and $b$ are real.
• We need to prove:
  $2x^2 = \left( a^2 + b^2 \right)^{\frac{1}{2}} + a$

Step 1: Square the complex number $z$

The complex number $z = x + iy$ can be squared as follows: $z^2 = (x + iy)^2 = x^2 + 2ixy - y^2.$ Separate the real and imaginary parts: $z^2 = (x^2 - y^2) + i(2xy).$ From the problem statement, $z^2 = a + ib$. Therefore, we can equate:

• Real part: $x^2 - y^2 = a$,
• Imaginary part: $2xy = b$.

Step 2: Express $x^2$ in terms of $a$ and $b$

We know:

1. $x^2 - y^2 = a$ (real part),
2. $2xy = b$ (imaginary part).

To isolate $x^2$, square the imaginary part equation: $(2xy)^2 = b^2 \implies 4x^2y^2 = b^2.$

From here, we write $x^2y^2$ as: $x^2y^2 = \frac{b^2}{4}.$

Step 3: Use a substitution for $x^2$ and $y^2$

To simplify, let $x^2 = p$ and $y^2 = q$. Then:

1. $p - q = a$ (from real part),
2. $pq = \frac{b^2}{4}$ (from squared imaginary part).

Now we solve for $p$ (or $x^2$) using these equations.

Step 4: Solving for $p$ (or $x^2$)

The two equations are:

1. $p - q = a$,
2. $pq = \frac{b^2}{4}$.

We solve this using algebraic substitution. From $p - q = a$, express $q$ as: $q = p - a.$ Substitute into $pq = \frac{b^2}{4}$: $p(p - a) = \frac{b^2}{4}.$ Simplify: $p^2 - ap = \frac{b^2}{4}.$ Rearrange into standard quadratic form: $p^2 - ap - \frac{b^2}{4} = 0.$

Step 5: Solve the quadratic equation

The quadratic equation is: $p^2 - ap - \frac{b^2}{4} = 0.$ Use the quadratic formula $p = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, where $A = 1$, $B = -a$, and $C = -\frac{b^2}{4}$: $p = \frac{-(-a) \pm \sqrt{(-a)^2 - 4(1)\left(-\frac{b^2}{4}\right)}}{2(1)}.$ Simplify: $p = \frac{a \pm \sqrt{a^2 + b^2}}{2}.$

Since $p = x^2 \geq 0$, we take the positive root: $x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}.$

Step 6: Verify the given equation

We need to prove: $2x^2 = \sqrt{a^2 + b^2} + a.$ Multiply $x^2$ by 2: $2x^2 = 2 \cdot \frac{a + \sqrt{a^2 + b^2}}{2}.$ Simplify: $2x^2 = a + \sqrt{a^2 + b^2}.$

Thus, the equation is proven.

Final Answer:

We have successfully proven: $2x^2 = \sqrt{a^2 + b^2} + a.$

Follow-up Questions:

1. What is the geometric interpretation of the result in terms of the modulus of a complex number?
2. How can you generalize this result for higher powers of $z$?
3. Can you explain how the quadratic formula applies here?
4. What is the relationship between $a$, $b$, and the magnitude of $z^2$?
5. How would the result change if $z$ were purely real or purely imaginary?

Tip:

When dealing with complex numbers, separating real and imaginary parts is often the key step for solving equations.