For any two complex numbers z_1 and z_2 and a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We use the property $|z|^2 = z \cdot \overline{z}$.$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (az_1 - bz_2)(\overline{az_1 - bz_2}) + (bz_1 + az_2)(\ove

Vedclass · Accepted Answer

$(a^2 + b^2)(|z_1|^2 + |z_2|^2)$

Vedclass · Answer

(B) We use the property $|z|^2 = z \cdot \overline{z}$.$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (az_1 - bz_2)(\overline{az_1 - bz_2}) + (bz_1 + az_2)(\overline{bz_1 + az_2})$$= (az_1 - bz_2)(a\overline{z_1} - b\overline{z_2}) + (bz_1 + az_2)(b\overline{z_1} + a\overline{z_2})$$= (a^2|z_1|^2 - ab z_1\overline{z_2} - ab \overline{z_1}z_2 + b^2|z_2|^2) + (b^2|z_1|^2 + ab z_1\overline{z_2} + ab \overline{z_1}z_2 + a^2|z_2|^2)$$= a^2|z_1|^2 + b^2|z_2|^2 + b^2|z_1|^2 + a^2|z_2|^2$$= (a^2 + b^2)|z_1|^2 + (a^2 + b^2)|z_2|^2$$= (a^2 + b^2)(|z_1|^2 + |z_2|^2)$

For any two complex numbers $z_1$ and $z_2$ and any real numbers $a$ and $b$,what is the value of $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$?

Similar Questions

The argument and modulus of $\frac{1 + i}{1 - i}$ are respectively:

If $x$ and $y$ are real numbers,which of the following is always true?

Let $\alpha$ be a fixed non-zero complex number with $|\alpha| < 1$ and $w = \frac{z-\alpha}{1-\bar{\alpha}z}$,where $z$ is a complex number. Then,

If $z = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6}$,then:

The modulus and amplitude of $\frac{1+2i}{1-(1-i)^{2}}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module