sqrt{(-3+4 i)(8+6 i)} = ? | vedclass

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Question

(C) Let $z = \sqrt{(-3+4 i)(8+6 i)}$.First,calculate the product inside the square root:$(-3+4 i)(8+6 i) = -24 - 18 i + 32 i + 24 i^2$Since $i^2 = -1$

Vedclass · Accepted Answer

$\pm(1+7 i)$

Vedclass · Answer

(C) Let $z = \sqrt{(-3+4 i)(8+6 i)}$.First,calculate the product inside the square root:$(-3+4 i)(8+6 i) = -24 - 18 i + 32 i + 24 i^2$Since $i^2 = -1$,we have:$-24 + 14 i - 24 = -48 + 14 i$.Now,we need to find $\sqrt{-48 + 14 i}$.Let $\sqrt{-48 + 14 i} = x + i y$,where $x, y \in \mathbb{R}$.Squaring both sides: $(x + i y)^2 = -48 + 14 i$$x^2 - y^2 + 2 i x y = -48 + 14 i$.Comparing real and imaginary parts:$x^2 - y^2 = -48$ $(1)$$2 x y = 14 \Rightarrow x y = 7$ $(2)$From $(2)$,$y = \frac{7}{x}$. Substituting into $(1)$:$x^2 - (\frac{7}{x})^2 = -48$$x^2 - \frac{49}{x^2} = -48$$x^4 + 48 x^2 - 49 = 0$$(x^2 + 49)(x^2 - 1) = 0$.Since $x \in \mathbb{R}$,$x^2 = 1$,so $x = \pm 1$.If $x = 1$,$y = 7$. If $x = -1$,$y = -7$.Thus,the square roots are $\pm(1 + 7 i)$.

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

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