To prove the given equation:

$\sqrt{a+bi} + \sqrt{a-bi} = \sqrt{2\left(a + \sqrt{a^2 + b^2}\right)},$

we'll proceed step by step.

Step 1: Represent complex square roots

The square root of a complex number $z = x + yi$ can be written in polar form or Cartesian form, but here we will directly use the formula for the square root of a complex number:

$\sqrt{a+bi} = \sqrt{\frac{r + a}{2}} + i \, \text{sgn}(b) \sqrt{\frac{r - a}{2}},$

where:

• $r = \sqrt{a^2 + b^2}$,
• $\text{sgn}(b)$ is the sign of $b$, which is $+1$ if $b > 0$ and $-1$ if $b < 0$.

Similarly, for $\sqrt{a-bi}$: $\sqrt{a-bi} = \sqrt{\frac{r + a}{2}} - i \, \sqrt{\frac{r - a}{2}}.$

Step 2: Add the two square roots

Now add $\sqrt{a+bi}$ and $\sqrt{a-bi}$:

$\sqrt{a+bi} + \sqrt{a-bi} = \left( \sqrt{\frac{r + a}{2}} + i \sqrt{\frac{r - a}{2}} \right) + \left( \sqrt{\frac{r + a}{2}} - i \sqrt{\frac{r - a}{2}} \right).$

The imaginary parts cancel out, leaving:

$\sqrt{a+bi} + \sqrt{a-bi} = 2 \sqrt{\frac{r + a}{2}}.$

Step 3: Simplify the result

Simplify $2 \sqrt{\frac{r + a}{2}}$:

$\sqrt{a+bi} + \sqrt{a-bi} = \sqrt{2} \cdot \sqrt{r + a}.$

Since $r = \sqrt{a^2 + b^2}$, this becomes:

$\sqrt{a+bi} + \sqrt{a-bi} = \sqrt{2 \left(a + \sqrt{a^2 + b^2}\right)}.$

Conclusion

Thus, we have proved the required equation:

$\sqrt{a+bi} + \sqrt{a-bi} = \sqrt{2\left(a + \sqrt{a^2 + b^2}\right)}.$

Do you want a deeper explanation or have any questions?

Related Questions

1. How can we represent a complex number in polar form?
2. What is the geometric interpretation of $\sqrt{a+bi}$?
3. How does this formula change if $b = 0$?
4. What are the real and imaginary parts of $\sqrt{a+bi}$?
5. Can you generalize this proof for $a+bi$ in polar coordinates?

Tip:

When working with square roots of complex numbers, try expressing them in polar form, as it often simplifies operations like addition or multiplication.