If Z = frac{(sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 | vedclass

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

Question

(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{2}}{|8 + 6i|^{2}}$.Calculate the modulus of each complex number:$|\sqrt{3} + i| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$.$|3i + 4| = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$.$|8 + 6i| = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$.Substitute these values into the expression for $|Z|$:$|Z| = \frac{2^{3} 	imes 5^{2}}{10^{2}} = \frac{8 	imes 25}{100} = \frac{200}{100} = 2$.Thus,$|Z| = 2$.

If $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$,then $|Z|$ is equal to

Similar Questions

If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$,then $({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) =$

Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

If $\log_{\tan 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} \right) < -2$,then:

Let $Z_1$ and $Z_2$ be any two complex numbers.
Statement $1: |Z_1 - Z_2| \ge |Z_1| - |Z_2|$
Statement $2: |Z_1 + Z_2| \le |Z_1| + |Z_2|$

If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4i, 3+i, 1-i$ and $2-3i$,then the correct relation among the following is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module