If sqrt{-3-4 i}=re^{i theta},then r^2 tan the | vedclass

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(D) Let $z = -3-4i$. We are given $\sqrt{z} = re^{i	heta}$.Squaring both sides,we get $z = r^2 e^{i2	heta} = r^2(\cos 2	heta + i \sin 2	heta)$.Com

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$-10$

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(D) Let $z = -3-4i$. We are given $\sqrt{z} = re^{i	heta}$.Squaring both sides,we get $z = r^2 e^{i2	heta} = r^2(\cos 2	heta + i \sin 2	heta)$.Comparing the real and imaginary parts with $z = -3-4i$:$r^2 \cos 2	heta = -3$ and $r^2 \sin 2	heta = -4$.We know that $r^2 = |z| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5$.Using the double angle formula for tangent: $	an 2	heta = \frac{r^2 \sin 2	heta}{r^2 \cos 2	heta} = \frac{-4}{-3} = \frac{4}{3}$.Using $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta} = \frac{4}{3}$,we get $6 	an 	heta = 4 - 4 	an^2 	heta$,or $4 	an^2 	heta + 6 	an 	heta - 4 = 0$.Dividing by $2$,$2 	an^2 	heta + 3 	an 	heta - 2 = 0$.Factoring gives $(2 	an 	heta - 1)(	an 	heta + 2) = 0$,so $	an 	heta = \frac{1}{2}$ or $	an 	heta = -2$.Since $z$ is in the third quadrant,its square root $\sqrt{z}$ lies in the second or fourth quadrant. For $\sqrt{z} = x+iy$,$xy = -2$,so $x$ and $y$ have opposite signs.Testing $	an 	heta = -2$,$r^2 	an 	heta = 5 	imes (-2) = -10$.

If $\sqrt{-3-4 i}=re^{i \theta}$,then $r^2 \tan \theta=$

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