Question

For any two non-zero complex numbers \(z_1\) and \(z_2\), if \(|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2\), then

• A. \( \text{Re}\left(\frac{z_1}{z_2}\right) = 0 \)
• B. \( \text{Im}\left(\frac{z_1}{z_2}\right) = 0 \)
• C. \( \text{Re}(z_1 z_2) = 0 \)
• D. \( \text{Im}(z_1 z_2) = 0 \)

Hint:

When dealing with magnitudes of complex numbers, the property \( |z|^2 = z \bar{z} \) is often very useful. The condition \( |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 \) implies that the complex numbers \(z_1\) and \(z_2\) are orthogonal if viewed as vectors in the complex plane, which means the angle between them is \( \frac{\pi}{2} \). This translates to their dot product being zero, which is \( \text{Re}(z_1 \bar{z_2}) = 0 \).

Answer 1

The Correct Option is A

The given condition is \(|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2\). We know that for any complex number \(z\), \(|z|^2 = z \bar{z}\). Therefore, we can write the given condition as: \( (z_1 + z_2)(\overline{z_1 + z_2}) = z_1 \bar{z_1} + z_2 \bar{z_2} \) \( (z_1 + z_2)(\bar{z_1} + \bar{z_2}) = z_1 \bar{z_1} + z_2 \bar{z_2} \) Expanding the left side: \( z_1 \bar{z_1} + z_1 \bar{z_2} + z_2 \bar{z_1} + z_2 \bar{z_2} = z_1 \bar{z_1} + z_2 \bar{z_2} \) Subtracting \(z_1 \bar{z_1} + z_2 \bar{z_2}\) from both sides: \( z_1 \bar{z_2} + z_2 \bar{z_1} = 0 \) We know that for any complex number \(w\), \(w + \bar{w} = 2 \text{Re}(w)\). Let \(w = z_1 \bar{z_2}\). Then \(\bar{w} = \overline{z_1 \bar{z_2}} = \bar{z_1} z_2\). So, \( z_1 \bar{z_2} + z_2 \bar{z_1} = 2 \text{Re}(z_1 \bar{z_2}) = 0 \) This implies \( \text{Re}(z_1 \bar{z_2}) = 0 \). Now, let's consider the options. We need to express this in terms of \( \frac{z_1}{z_2} \). We can write \( z_1 \bar{z_2} \) as \( z_1 \bar{z_2} = \frac{z_1 \bar{z_2}}{|z_2|^2} |z_2|^2 = \frac{z_1}{z_2} |z_2|^2 \). So, \( \text{Re}\left(\frac{z_1}{z_2} |z_2|^2\right) = 0 \). Since \( |z_2|^2 \) is a real non-zero number (as \(z_2\) is a non-zero complex number), we can divide by it. Therefore, \( |z_2|^2 \text{Re}\left(\frac{z_1}{z_2}\right) = 0 \) This implies \( \text{Re}\left(\frac{z_1}{z_2}\right) = 0 \). Alternatively, let \(z_1 = x_1 + i y_1\) and \(z_2 = x_2 + i y_2\). \( |z_1 + z_2|^2 = (x_1+x_2)^2 + (y_1+y_2)^2 = x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 \) \( |z_1|^2 + |z_2|^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 \) Given \(|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2\), we have: \( x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 \) This simplifies to: \( 2x_1x_2 + 2y_1y_2 = 0 \) \( x_1x_2 + y_1y_2 = 0 \) Now let's look at \( \frac{z_1}{z_2} \): \( \frac{z_1}{z_2} = \frac{x_1 + i y_1}{x_2 + i y_2} = \frac{(x_1 + i y_1)(x_2 - i y_2)}{(x_2 + i y_2)(x_2 - i y_2)} = \frac{x_1x_2 - i x_1y_2 + i y_1x_2 + y_1y_2}{x_2^2 + y_2^2} \) \( \frac{z_1}{z_2} = \frac{(x_1x_2 + y_1y_2) + i(y_1x_2 - x_1y_2)}{x_2^2 + y_2^2} \) The real part of \( \frac{z_1}{z_2} \) is \( \text{Re}\left(\frac{z_1}{z_2}\right) = \frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} \). Since we found \( x_1x_2 + y_1y_2 = 0 \) and \( x_2^2 + y_2^2 \neq 0 \) (as \(z_2\) is non-zero), we have: \( \text{Re}\left(\frac{z_1}{z_2}\right) = \frac{0}{x_2^2 + y_2^2} = 0 \).