Let z 2

Let z = 2 - sqrt(3i). Find z⁶ and express your answer in rectangular form. if z = 2 - 2sqrt(3 i) then r = |z| = sqrt(2 ^ 2 + (- 2sqrt(3)) ^ 2) = sqrt(16) = 4 and theta = tan -2√3/2=-π/3

Correct answer:

x =  8.999
y =  2.2439

Step-by-step explanation:

$$\begin{gathered} \sqrt{3i}=x_1+y_1\cdot i \\ {x}_1=\sqrt{3}/\sqrt{2}\doteq 1.2247 \\ {y}_1=x_1=1.2247\doteq 1.2247 \\ z=x_2+i\cdot y_2 \\ {x}_2=2-x_1=2-1.2247\doteq 0.7753 \\ {y}_2=0-y_1=0-1.2247\doteq -1.2247 \\ r=|z| \\ r=\sqrt{x_2^2+y_2^2}=\sqrt{0.7753^2+(-1.2247{)}^2}\doteq 1.4495 \\ \theta =\arctan (y_2/x_2)=\arctan ((-1.2247)/0.7753)\doteq -1.0065\text{rad} \\ {z}^6=r^6\cdot (\cos 6\theta +i\sin 6\theta ) \\ {r}_2=r^6=1.4495^6\doteq 9.2745 \\ {\theta }_2=6\cdot \theta =6\cdot (-1.0065)\doteq -6.0388\text{rad} \\ x=r_2\cdot \cos ({\theta }_2)=9.2745\cdot \cos ((-6.0388))=8.999 \end{gathered}$$

$y=r_2\cdot \sin ({\theta }_2)=9.2745\cdot \sin ((-6.0388))=2.2439$

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