Question

please help
For z = 2cis 45°, find z3 in rectangular form

Answer 1

Is that suppose to be \[z^3 \]

Answer 2

yes

Answer 3

In polar form:
\[z = re^{i \theta} = rcos(\theta) + irsin(\theta) \\z = 2e^{\frac{\pi}{4}i} \\ 0 \leq \theta < 2\pi\]
In general, its way easier to deal with powers in polar, form, just be aware that this is not a unique description of the point, but we will restrict our angle as shown above. So then:
\[z^3 = r^3e^{3\theta i} \\ z^3 = 8e^{\frac{3 \pi}{4}i}\]

Then to convert to rectangular, use Euler's eqn:
\[z^3 = 8e^{\frac{3 \pi}{4}i}\ = 8\cos(\frac{3 \pi}{4})+i8\sin (\frac{3 \pi}{4}) \\ z^3 = -8\frac{\sqrt{2}}{2}+8\frac{\sqrt{2}}{2}i=-4\sqrt{2}+4\sqrt{2}i \]

Answer 4

ok thank you so much!