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On Khan academy their is this question asking:

Find the solution of the equation whose argument is between 180 and 270 degrees, z^5 = -243i

Now their is also a document on the khan academy site that says if their is a negative in the equation, such as the -243i part, that the argument will be 5(theta) = 180 + k(360)... And if their is no negative then the argument will be 5(theta) = 90 + k(360)... However when I click to get a hint for the question it says that the argument is 5(theta) = 270 + k(360). This is really confusing me because on a similar question that asks for the solution whose argument is between 225 degrees and 315 degrees, when I ask for a hint, they tell me the argument is 5(theta) = 90 + k(360).

Can someone explain how they came to the argument of 270?

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Observe that $i^5 = i$ and $(-3)^5 = -243$, so one solution is $-3i = -3e^{\pi i/2}$, which is the same as $3e^{3\pi i/2}$, and $3\pi/2$ radians is $270$ degrees.

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  • $\begingroup$ How did you go from -3i = -3e^(pi/2)i to 3e^(3pi/2)i ? $\endgroup$ – thankgodforthissite Nov 26 '15 at 23:05
  • $\begingroup$ We have Euler's formula, $$re^{i\theta} = r(\cos\theta + i\sin\theta).$$ So $$-3e^{\pi i/2} = -3(\cos(\pi/2) + i\sin(\pi/2) = -3i.$$ And changing from $\pi/2 \to 3\pi/2$ is the same as rotating by $\pi$, i.e. multiplying by $-1$. $\endgroup$ – Ethan Alwaise Nov 27 '15 at 0:28
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HINT:

$$z^5=-243i\Longleftrightarrow$$ $$z^5=\left|-243i\right|e^{\arg\left(-243i\right)i}\Longleftrightarrow$$ $$z^5=243e^{-\frac{\pi}{2}i}\Longleftrightarrow$$ $$z=\left(243e^{\left(2\pi k-\frac{\pi}{2}\right)i}\right)^{\frac{1}{5}}\Longleftrightarrow$$ $$z=\sqrt[5]{243}e^{\frac{1}{5}\left(2\pi k-\frac{\pi}{2}\right)i}\Longleftrightarrow$$ $$z=3e^{\frac{\pi\left(4k-1\right)}{10}i}$$

With $k\in\mathbb{Z}$ and $k:0-4$.

So the solutions are:

$$z_0=3e^{\frac{\pi\left(4\cdot 0-1\right)}{10}i}=3e^{-\frac{\pi i}{10}}$$ $$z_1=3e^{\frac{\pi\left(4\cdot 1-1\right)}{10}i}=3e^{\frac{3\pi i}{10}}$$ $$z_2=3e^{\frac{\pi\left(4\cdot 2-1\right)}{10}i}=3e^{\frac{7\pi i}{10}}$$ $$z_3=3e^{\frac{\pi\left(4\cdot 3-1\right)}{10}i}=3e^{-\frac{9\pi i}{10}}$$ $$z_4=3e^{\frac{\pi\left(4\cdot 4-1\right)}{10}i}=3e^{-\frac{\pi i}{2}}=-3i$$


Now filter the ones you need!

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