One of the values of sqrt{24-70 i}+sqrt{-24+7 | vedclass

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

Question

(D) Let $z_1 = 24-70 i$ and $z_2 = -24+70 i$.We need to find $\sqrt{z_1} + \sqrt{z_2}$.Note that $z_2 = -z_1$,so we are looking for $\sqrt{z_1} + \sqr

Vedclass · Accepted Answer

$-12-2 i$

Vedclass · Answer

(D) Let $z_1 = 24-70 i$ and $z_2 = -24+70 i$.We need to find $\sqrt{z_1} + \sqrt{z_2}$.Note that $z_2 = -z_1$,so we are looking for $\sqrt{z_1} + \sqrt{-z_1}$.Let $\sqrt{24-70 i} = x+iy$. Squaring both sides,$x^2-y^2+2ixy = 24-70 i$.Equating real and imaginary parts: $x^2-y^2 = 24$ and $2xy = -70 \implies xy = -35$.Since $x^2+y^2 = \sqrt{24^2+(-70)^2} = \sqrt{576+4900} = \sqrt{5476} = 74$.Adding $x^2-y^2=24$ and $x^2+y^2=74$,we get $2x^2 = 98 \implies x^2 = 49 \implies x = \pm 7$.If $x=7, y=-5$. If $x=-7, y=5$. So $\sqrt{24-70 i} = \pm(7-5 i)$.Similarly,$\sqrt{-24+70 i} = \pm(5-7 i)$.Possible sums are:$1$) $(7-5 i) + (5-7 i) = 12-12 i$$2$) $(7-5 i) - (5-7 i) = 2+2 i$$3$) $-(7-5 i) + (5-7 i) = -2-2 i$$4$) $-(7-5 i) - (5-7 i) = -12+12 i$Wait,checking the options provided,there might be a typo in the question or options. Re-evaluating: $\sqrt{24-70 i} = \pm(7-5 i)$ and $\sqrt{-24+70 i} = \pm i(7-5 i) = \pm(5+7 i)$.Sum: $\pm(7-5 i) \pm(5+7 i)$.Possible values: $(7-5 i) + (5+7 i) = 12+2 i$,$(7-5 i) - (5+7 i) = 2-12 i$,$-(7-5 i) + (5+7 i) = -2+12 i$,$-(7-5 i) - (5+7 i) = -12-2 i$.Option $D$ matches $-12-2 i$.

One of the values of $\sqrt{24-70 i}+\sqrt{-24+70 i}$ is

Similar Questions

If $(-7 - 24i)^{1/2} = x - iy$,then $x^2 + y^2 = $

If $\tanh^{-1}(x+iy) = \frac{1}{2} \tanh^{-1}\left(\frac{2x}{1+x^2+y^2}\right) + \frac{i}{2} \tan^{-1}\left(\frac{2y}{1-x^2-y^2}\right)$,where $x, y \in \mathbb{R}$,then $\tanh^{-1}(iy) =$

$\sqrt[4]{17 + 12\sqrt{2}} = $

If the real parts of $\sqrt{-5-12 i}$ and $\sqrt{5+12 i}$ are positive,the real part of $\sqrt{-8-6 i}$ is negative,and $a+i b = \frac{\sqrt{-5-12 i}+\sqrt{5+12 i}}{\sqrt{-8-6 i}}$,then $2 a+b =$

$\cosh (\log 4) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module