kramtus51

Answered

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.)

Answer & Explanation

jgardner33v4

Expert

2022-01-19Added 35 answers

${\left(\sqrt{13}-i\right)}^{4}$
${\left(\sqrt{13}-i\right)}^{2}{\left(\sqrt{13}-i\right)}^{2}$
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

2022-01-20Added 36 answers

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

2022-01-24Added 366 answers

$\left(\sqrt{13}-i{\right)}^{2}\left(\sqrt{13}-i{\right)}^{2}$ 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: $=r{e}^{i\theta }⇒4\sqrt{542}{e}^{i{\mathrm{tan}}^{-1}\left(-\frac{\sqrt{13}}{23}\right)}$

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