Q.For any two complex numbers $z_1, z_2$ and any real numbers $a, b$, $\;|az_1-bz_2|^2+|bz_1+az_2|^2=\;$ _____.

The key idea here is the property of the modulus of a complex number: $|z|^2 = z\bar{z}$.

We expand each term using this property. Since $a$ and $b$ are real numbers, their conjugates are themselves ($\bar{a}=a, \bar{b}=b$).

1. Expand the first term:

   $|az_1-bz_2|^2 = (az_1-bz_2)(\overline{az_1-bz_2}) = (az_1-bz_2)(a\bar{z_1}-b\bar{z_2})$

   $= a^2 z_1\bar{z_1} - ab z_1\bar{z_2} - ab z_2\bar{z_1} + b^2 z_2\bar{z_2}$

   $= a^2|z_1|^2 - ab(z_1\bar{z_2} + z_2\bar{z_1}) + b^2|z_2|^2$

2. Expand the second term:

   $|bz_1+az_2|^2 = (bz_1+az_2)(\overline{bz_1+az_2}) = (bz_1+az_2)(b\bar{z_1}+a\bar{z_2})$

   $= b^2 z_1\bar{z_1} + ab z_1\bar{z_2} + ab z_2\bar{z_1} + a^2 z_2\bar{z_2}$

   $= b^2|z_1|^2 + ab(z_1\bar{z_2} + z_2\bar{z_1}) + a^2|z_2|^2$

3. Add the two expanded expressions: …

When dealing with expressions involving the square of the magnitude of a complex number, such as $|z|^2$, the most efficient approach is almost always to use the identity $z\bar{z} = |z|^2$. This identity is incredibly powerful because it transforms a geometric concept (magnitude) into an algebraic product, allowing us to use standard algebraic manipulation.

Let's quickly recall why this identity holds. If $z = x + iy$, then $\bar{z} = x - iy$.

So, $z\bar{z} = (x+iy)(x-iy) = x^2 - (iy)^2 = x^2 - i^2y^2 = x^2 - (-1)y^2 = x^2+y^2$.

We also know that $|z| = \sqrt{x^2+y^2}$, so $|z|^2 = x^2+y^2$.

Thus, $z\bar{z} = |z|^2$.

This approach avoids the need to substitute $z_1 = x_1+iy_1$ and $z_2 = x_2+iy_2$ and then expand everything, which would be much more tedious and error-prone. The key is to remember the properties of complex conjugates:

• $\overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2}$
• $\overline{z_1 z_2} = \bar{z_1} \bar{z_2}$
• $\overline{kz} = k\bar{z}$ if $k$ is a real number. This last property is vital here because $a$ and $b$ are real numbers.

Let's apply this to the given expression.

1. Expand the first term, $|az_1-bz_2|^2$: Using the identity $|z|^2 = z\bar{z}$, we can write:

Now, apply the conjugate properties: $\overline{(az_1-bz_2)} = \overline{az_1} - \overline{bz_2} = a\bar{z_1} - b\bar{z_2}$ (since $a, b$ are real).
So, the expression becomes:

Expand this product:

Substitute $z_1\bar{z_1} = |z_1|^2$ and $z_2\bar{z_2} = |z_2|^2$:

2. Expand the second term, $|bz_1+az_2|^2$: Similarly, using $|z|^2 = z\bar{z}$:

Apply conjugate properties: $\overline{(bz_1+az_2)} = \overline{bz_1} + \overline{az_2} = b\bar{z_1} + a\bar{z_2}$.
So, the expression becomes:

Expand this product:

Substitute $z_1\bar{z_1} = |z_1|^2$ and $z_2\bar{z_2} = |z_2|^2$:

3. Add the expanded terms (*) and (): …