kramtus51

2022-01-18

Write the expression in rectangular form, $x+yi$, and in exponential form $r{e}^{i\theta }$.
${\left(\sqrt{13}-i\right)}^{4}$
The rectangular form of the given expression is ?, and the exponential form of the given expression is ? (Simplify your answers. Use integers or decimals for any numbers in the expressions. Round the final answer to three decimal places as needed intermediate values to four decimal places as needed.)

jgardner33v4

Expert

${\left(\sqrt{13}-i\right)}^{4}$
${\left(\sqrt{13}-i\right)}^{2}{\left(\sqrt{13}-i\right)}^{2}$
Now ${\left(A-B\right)}^{2}={A}^{2}+{B}^{2}-2AB$
$\left(13+{i}^{2}-2\sqrt{13}i\right)\left(13+{i}^{2}-2\sqrt{13}i\right)$
${i}^{2}=-1$
$\left(12-2\sqrt{13}i\right)\left(12-2\sqrt{13}i\right)$
$\left(144+4×13{i}^{-2}-4\sqrt{13}i\right)$
$\left(144-52-4\sqrt{13}i\right)$ $=92-4\sqrt{13}i\to$ rectangular form
$r=\sqrt{{\left(92\right)}^{2}+{\left(4\sqrt{13}\right)}^{2}}$
$r=\sqrt{8464+208}⇒r\sqrt{8672}$
$r=4\sqrt{542}$
$\theta ={\mathrm{tan}}^{4}\left(-\frac{4\sqrt{13}}{92}\right)$
$\theta ={\mathrm{tan}}^{-1}\left(-\frac{\sqrt{13}}{23}\right)$ $=r{e}^{i\theta }⇒4\sqrt{542}{e}^{i{\mathrm{tan}}^{-1}\left(-\frac{\sqrt{13}}{23}\right)}$

reinosodairyshm

Expert

Say ${\left(\sqrt{13}-i\right)}^{4}$
$⇒{\left({\left(\sqrt{13}-1\right)}^{2}\right)}^{2}$
$⇒{\left({\left(\sqrt{13}\right)}^{2}+{\left(i\right)}^{2}-2\cdot \sqrt{13}.i\right)}^{2}$
$⇒{\left(13+\left(-1\right)-2\sqrt{13}i\right)}^{2}$
$⇒{\left(12-2\sqrt{13}i\right)}^{2}$
$⇒{\left(12\right)}^{2}+{\left(2\sqrt{13}i\right)}^{2}-2\cdot 12\cdot 2\sqrt{13}i$
$⇒144+52{i}^{2}-48\sqrt{13}i$
$⇒144-52-48\sqrt{13}i$
$⇒92+\left(-48\sqrt{13}i\right)$
This is in the form of $x+iy$$x=92,y=\left(-48\sqrt{13}\right)$
if $r{e}^{i\theta }$
$v=\sqrt{{\left(92\right)}^{2}+{\left(-48\sqrt{13}\right)}^{2}}$
$=\sqrt{{x}^{2}+{y}^{n}}$
$⇒r=\sqrt{38,416}$
$⇒r=196$
$⇒\theta =\left({360}^{\circ }-{61.99}^{\circ }\right)$
$⇒\theta ={298.01}^{\circ }$
$⇒\theta \approx {298}^{\circ }=\frac{77}{45}$ $=r{e}^{i\theta }=196{e}^{i\cdot {298}^{\circ }}$$=196\cdot {e}^{i\frac{77}{45}}$ (answer)

alenahelenash

Expert

$\left(\sqrt{13}-i{\right)}^{2}\left(\sqrt{13}-i{\right)}^{2}$ Now $\left(A-B{\right)}^{2}={A}^{2}+{B}^{2}-2AB$ $\left(13+{i}^{2}-2\sqrt{13}i\right)\left(13+{i}^{2}-2\sqrt{13}i\right)$ ${i}^{2}=-1$ $\left(12-2\sqrt{13}i\right)\left(12-2\sqrt{13}i\right)$ $\left(144+4×13{i}^{-2}-4\sqrt{13}i\right)$ $\left(144-52-4\sqrt{13}i\right)$ $=92-4\sqrt{13}i\to$ rectangular form Now $r=\sqrt{\left(92{\right)}^{2}+\left(4\sqrt{13}{\right)}^{2}}$ $r=\sqrt{8464+208}⇒r\sqrt{8672}$ $r=4\sqrt{542}$ $\theta ={\mathrm{tan}}^{4}\left(-\frac{4\sqrt{13}}{92}\right)$ $\theta ={\mathrm{tan}}^{-1}\left(-\frac{\sqrt{13}}{23}\right)$ Answer: $=r{e}^{i\theta }⇒4\sqrt{542}{e}^{i{\mathrm{tan}}^{-1}\left(-\frac{\sqrt{13}}{23}\right)}$