If z is a complex number, then:
- A$|\text{z}^2|>|\text{z}|^2$
- B$|\text{z}^2|=|\text{z}|^2$
- C$|\text{z}^2|<|\text{z}|^2$
- D$|\text{z}^2|\geq|\text{z}|^2$
Solution:
Let z = x + yi
|z| = |x + yi| and |z|2 = |x + yi|2
⇒ |z|2 = x2 + y2 .....(i)
Now, z2 = x2 + y2i2 + 2xyi
z2 = x2 - y2 + 2xyi
$|\text{z}^2|=\sqrt{(\text{x}^2-\text{y}^2)^2+(2\text{xy})^2}$
$=\sqrt{\text{x}^4+\text{y}^4-2\text{x}^2\text{y}^2+4\text{x}^2\text{y}^2}$
$=\sqrt{\text{x}^4+\text{y}^4+2\text{x}^2\text{y}^2}$
$=\sqrt{(\text{x}^2+\text{y}^2)^2}$
So, $|\text{z}|^2=\text{x}^2+\text{y}^2=|\text{z}|^2$
So, $|\text{z}|^2=|\text{z}^2|$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
A plane is parallel to yz-plane so it is perpendicular to:
If sets A and B are defined as $\text{A}=\Big\{(\text{x},\text{y})|\text{y}=\frac{1}{\text{x}},0\neq\text{x}\in\text{R}\Big\}\ \text{B}=\{(\text{x},\text{y})|\text{y}=-\text{x},\text{x}\in\text{R}\},$ then
$\text{A}\cap\text{B}=\text{A}$
$\text{A}\cap\text{B}=\text{B}$
$\text{A}\cap\text{B}=\phi$
$\text{A}\cup\text{B}=\text{A}$
A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected randomwise, the probability that it is black or red ball is:
If -40°F is equal to -40°C and 0°C is equal to 32°F then find the value of 40°C: