If (x+iy)^{frac{1}{3}} = 5+3i,then 3x+5y = | vedclass

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

Question

(D) Given $(x+iy)^{\frac{1}{3}} = 5+3i$. Cubing both sides,we get $x+iy = (5+3i)^3$. Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have:

Vedclass · Accepted Answer

$960$

Vedclass · Answer

(D) Given $(x+iy)^{\frac{1}{3}} = 5+3i$. Cubing both sides,we get $x+iy = (5+3i)^3$. Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have: $x+iy = 5^3 + 3(5^2)(3i) + 3(5)(3i)^2 + (3i)^3$. $x+iy = 125 + 3(25)(3i) + 15(9i^2) + 27i^3$. Since $i^2 = -1$ and $i^3 = -i$,we get: $x+iy = 125 + 225i + 135(-1) + 27(-i)$. $x+iy = 125 + 225i - 135 - 27i$. $x+iy = (125-135) + (225-27)i$. $x+iy = -10 + 198i$. Comparing real and imaginary parts,$x = -10$ and $y = 198$. Now,calculate $3x+5y = 3(-10) + 5(198) = -30 + 990 = 960$.

If $(x+iy)^{\frac{1}{3}} = 5+3i$,then $3x+5y = $

Similar Questions

If $\frac{3+2i \sin \theta}{1-2i \sin \theta}$ is a real number and $0 < \theta < 2\pi$,then $\theta$ is equal to

Let $A = \left\{ \theta \in \left( -\frac{\pi}{2}, \pi \right) : \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \text{ is purely imaginary} \right\}$. Then the sum of the elements in $A$ is

The value of $\frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}}$,where $i=\sqrt{-1}$,is

Express $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$ in the form of $a+ib$.

The value of $i^{1 + 3 + 5 + ... + (2n + 1)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module