The square root of 7+24 i is: | vedclass

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

Question

(A) Let $\sqrt{7+24 i} = x+iy$. Squaring both sides,we get $7+24 i = (x^2-y^2) + 2xyi$. Equating the real and imaginary parts,we have $x^2-y^2 = 7$ an

Vedclass · Accepted Answer

$4+3 i$

Vedclass · Answer

(A) Let $\sqrt{7+24 i} = x+iy$. Squaring both sides,we get $7+24 i = (x^2-y^2) + 2xyi$. Equating the real and imaginary parts,we have $x^2-y^2 = 7$ and $2xy = 24$,which implies $xy = 12$ or $y = \frac{12}{x}$. Substituting $y$ into the first equation: $x^2 - (\frac{12}{x})^2 = 7 \Rightarrow x^2 - \frac{144}{x^2} = 7$. Multiplying by $x^2$: $x^4 - 7x^2 - 144 = 0$. Let $u = x^2$,then $u^2 - 7u - 144 = 0$. Factoring gives $(u-16)(u+9) = 0$. Since $x$ is real,$x^2 = 16$,so $x = \pm 4$. If $x = 4$,$y = 3$. If $x = -4$,$y = -3$. Thus,the square roots are $\pm(4+3i)$.

The square root of $7+24 i$ is:

Similar Questions

$\sqrt{(-3+4 i)(8+6 i)} = ?$

If $2 \cosh 2x + 10 \sinh 2x = 5$,then $x =$

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

$\sqrt[3]{61 - 46\sqrt{5}} = $

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module