If x+iy = frac{(3+2i)(4-7i)(12+13i)}{(13-12i) | vedclass

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

Question

(C) Given,$z = x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$.Taking the modulus on both sides,we have $|z| = |x+iy| = \sqrt{x^2+y^2}$.Usi

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Given,$z = x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$.Taking the modulus on both sides,we have $|z| = |x+iy| = \sqrt{x^2+y^2}$.Using the property $|\frac{z_1 z_2 z_3}{z_4 z_5 z_6}| = \frac{|z_1| |z_2| |z_3|}{|z_4| |z_5| |z_6|}$,we get:$|z| = \frac{|3+2i| \cdot |4-7i| \cdot |12+13i|}{|13-12i| \cdot |2-3i| \cdot |11+3i|}$.Note that $|3+2i| = |2-3i| = \sqrt{3^2+2^2} = \sqrt{13}$.Also,$|12+13i| = |13-12i| = \sqrt{12^2+13^2} = \sqrt{313}$.Substituting these values,we get:$|z| = \frac{\sqrt{13} \cdot |4-7i| \cdot \sqrt{313}}{\sqrt{313} \cdot \sqrt{13} \cdot |11+3i|} = \frac{|4-7i|}{|11+3i|}$.$|z| = \frac{\sqrt{4^2+(-7)^2}}{\sqrt{11^2+3^2}} = \frac{\sqrt{16+49}}{\sqrt{121+9}} = \frac{\sqrt{65}}{\sqrt{130}} = \frac{\sqrt{65}}{\sqrt{2 	imes 65}} = \frac{1}{\sqrt{2}}$.Since $|z| = \sqrt{x^2+y^2}$,we have $\sqrt{x^2+y^2} = \frac{1}{\sqrt{2}}$.Squaring both sides,$x^2+y^2 = \frac{1}{2}$.

If $x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$,then $x^2+y^2=$

Similar Questions

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

If $a > 0$ and $z = \frac{(1 + i)^2}{a - i}$,where $i = \sqrt{-1}$,has magnitude $\sqrt{\frac{2}{5}}$,then $\overline{z}$ is equal to

If $\frac{(p + i)^2}{2p - i} = \mu + i\lambda$,then $\mu^2 + \lambda^2$ is equal to

Let $z$ be a complex number such that $|z| + z = 3 + i$,where $i = \sqrt{-1}$. Then $|z| = $

If $\left|z+\frac{2}{z}\right|=2$,then the maximum value of $|z|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module