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Express $z=4\sqrt2(1+i)$ in modulus/argument form. Hence find the two square roots of $z$ and mark their representations on an Argand Diagram.

So far I've worked out the mod/arg form of the complex number which is just

$$ z = 8(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) \\ $$

Then used $\alpha = \frac{\theta +360k}{n}$ where $k$ is $0,1,2,...,n-1$ and got two results for the argument. The first being $\frac{-7\pi}{8}$ and the other being $\frac{\pi}{8}$.

My final answers were

$$ z_1 = 2\sqrt2(\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) \\ z_2 = 2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$

For some reason, the answers from the book were

$$ z_1 = 2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) \\ z_2 = -2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$

What was it that I did wrong in my calculations? Thanks in advance!

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Both are correct. In fact, they are the same.

$$ 2\sqrt 2 (\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) = -2 \sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$

Moreover, $-2\sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})$ looks better.

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  • $\begingroup$ I see, thank you! At least I know I'm doing it right! Thanks again! @victor $\endgroup$ – S.E. Chahine Oct 23 '14 at 11:54

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