If the real parts of sqrt{-5-12 i} and sqrt{5 | vedclass

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(C) Let $\sqrt{-5-12 i} = x+y i$,where $x > 0$. $-5-12 i = (x+y i)^2 = x^2-y^2+2 x y i$. Equating real and imaginary parts: $x^2-y^2 = -5$ and $2 x y

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$-3$

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(C) Let $\sqrt{-5-12 i} = x+y i$,where $x > 0$. $-5-12 i = (x+y i)^2 = x^2-y^2+2 x y i$. Equating real and imaginary parts: $x^2-y^2 = -5$ and $2 x y = -12$. $x^2+y^2 = \sqrt{(-5)^2+(-12)^2} = \sqrt{25+144} = 13$. Adding the equations $x^2-y^2 = -5$ and $x^2+y^2 = 13$ gives $2 x^2 = 8$,so $x = 2$ (since $x > 0$). Then $y = -3$. Thus,$\sqrt{-5-12 i} = 2-3 i$. Similarly,for $\sqrt{5+12 i} = u+v i$ with $u > 0$: $u^2-v^2 = 5$ and $2 u v = 12$. $u^2+v^2 = 13$. Adding gives $2 u^2 = 18$,so $u = 3$ and $v = 2$. Thus,$\sqrt{5+12 i} = 3+2 i$. For $\sqrt{-8-6 i} = m+n i$ with $m $m^2+n^2 = \sqrt{(-8)^2+(-6)^2} = 10$. Adding gives $2 m^2 = 2$,so $m = -1$ (since $m Then $n = 3$. Thus,$\sqrt{-8-6 i} = -1+3 i$. Now,$a+b i = \frac{(2-3 i)+(3+2 i)}{-1+3 i} = \frac{5-i}{-1+3 i}$. Multiplying by the conjugate: $\frac{5-i}{-1+3 i} 	imes \frac{-1-3 i}{-1-3 i} = \frac{-5-15 i+i+3 i^2}{1+9} = \frac{-8-14 i}{10} = -0.8-1.4 i$. So $a = -0.8$ and $b = -1.4$. $2 a+b = 2(-0.8)+(-1.4) = -1.6-1.4 = -3$.

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 $2 a+b =$

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