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If $\sqrt{-3-4 i}=\mathrm{re}^{\mathrm{i} \theta}$, then $\mathrm{r}^2 \tan \theta=$
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Verified Answer
The correct answer is:
$-5$
Given : $\sqrt{-3-4 i}=r e$ ...(i)
Let $\sqrt{-3-4 i}=x+i y$ ...(ii)
$\Rightarrow \quad-3-4 i=(x+i y)^2=x^2-y^2+2 i x y$
Comparing both sides,
$x^2-y^2=-3$
$\Rightarrow 2 x y=-4 \Rightarrow x y=-2$
Solving above equations, we get
$x= \pm 1$ and $y=\mp 2$
So, we have
$\sqrt{-3-4 i}=(-1+2 i),(1-2 i)$
Now, $-1+2 i=z_1$ (Let)
$\begin{aligned} & \Rightarrow \quad\left|z_1\right|=\sqrt{5}=r \\ & \Rightarrow \quad \arg \left(z_1\right)=\pi-\tan ^{-1}\left(\frac{2}{1}\right)=\pi-\tan ^{-1} 2 \\ & \Rightarrow \quad \theta=\pi-\tan ^{-1} 2 \\ & \Rightarrow \quad 2=\tan (\pi-\theta) \\ & \Rightarrow \quad-\tan \theta=2 \Rightarrow \tan \theta=-2\end{aligned}$
Now, $r^2 \tan \theta=(5) \times(-2)=-10$.
Let $\sqrt{-3-4 i}=x+i y$ ...(ii)
$\Rightarrow \quad-3-4 i=(x+i y)^2=x^2-y^2+2 i x y$
Comparing both sides,
$x^2-y^2=-3$
$\Rightarrow 2 x y=-4 \Rightarrow x y=-2$
Solving above equations, we get
$x= \pm 1$ and $y=\mp 2$
So, we have
$\sqrt{-3-4 i}=(-1+2 i),(1-2 i)$
Now, $-1+2 i=z_1$ (Let)
$\begin{aligned} & \Rightarrow \quad\left|z_1\right|=\sqrt{5}=r \\ & \Rightarrow \quad \arg \left(z_1\right)=\pi-\tan ^{-1}\left(\frac{2}{1}\right)=\pi-\tan ^{-1} 2 \\ & \Rightarrow \quad \theta=\pi-\tan ^{-1} 2 \\ & \Rightarrow \quad 2=\tan (\pi-\theta) \\ & \Rightarrow \quad-\tan \theta=2 \Rightarrow \tan \theta=-2\end{aligned}$
Now, $r^2 \tan \theta=(5) \times(-2)=-10$.
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