- Thread starter
- #1

#### sweatingbear

##### Member

- May 3, 2013

- 91

How many real and non-real roots does \(\displaystyle z^5 = 32\) have? \(\displaystyle z^9 = -4\)?

For \(\displaystyle z^5 = 32\): \(\displaystyle z^5 = r^5 ( \cos 5v + i \sin 5v )\) and \(\displaystyle 32 = 32 ( \cos 0 + i \sin 0 )\) yields

\(\displaystyle r = 2 \\ 5v = n \cdot 2\pi \iff v = n \cdot \dfrac {2\pi}5\)

So all roots are given by

\(\displaystyle z = 2 \left( \cos \left( n \cdot \frac {2\pi}5 \right) + i \sin \left( n \cdot \frac {2\pi}5 \right) \right)\)

where \(\displaystyle 0 \leqslant n \leqslant 4, \ n \in \mathbb{Z}\). Let us rule out all real roots by letting the imaginary part equal zero.

\(\displaystyle \sin \left( n \cdot \frac {2\pi}5 \right) = 0 \iff n \cdot \frac {2\pi}5 = k \cdot \pi \iff n = \frac {5k}2 \)

In order for \(\displaystyle n\) to be an integer, \(\displaystyle 2\) must divide \(\displaystyle k\). Thus \(\displaystyle k = 2p\) where \(\displaystyle p \in \mathbb{Z}\) and consequently \(\displaystyle n = 5p\). With \(\displaystyle 0 \leqslant n \leqslant 4\) we have

\(\displaystyle 0 \leqslant 5p \leqslant 4 \iff 0 \leqslant p \leqslant 0.8 \implies p \in \{ 0 \}\)

So there is one real root and the remaining four are non-real. Similar arguments for \(\displaystyle z^9 = -4\) yield

\(\displaystyle 9v = \pi + n \cdot 2\pi \iff v = \frac {\pi}9 + n \cdot \frac {2\pi}9\)

Equating the imaginary part to zero yields

\(\displaystyle \frac {\pi}9 + n \cdot \frac {2\pi}9 = k \cdot \pi \iff n = \frac {9k-1}2\)

\(\displaystyle 2\) must divide \(\displaystyle 9k-1\) so \(\displaystyle 9k-1 = 2p\). Using \(\displaystyle 0 \leqslant n \leqslant 8\) however yields \(\displaystyle 0 \leqslant p \leqslant 8\). The equation does

For \(\displaystyle z^5 = 32\): \(\displaystyle z^5 = r^5 ( \cos 5v + i \sin 5v )\) and \(\displaystyle 32 = 32 ( \cos 0 + i \sin 0 )\) yields

\(\displaystyle r = 2 \\ 5v = n \cdot 2\pi \iff v = n \cdot \dfrac {2\pi}5\)

So all roots are given by

\(\displaystyle z = 2 \left( \cos \left( n \cdot \frac {2\pi}5 \right) + i \sin \left( n \cdot \frac {2\pi}5 \right) \right)\)

where \(\displaystyle 0 \leqslant n \leqslant 4, \ n \in \mathbb{Z}\). Let us rule out all real roots by letting the imaginary part equal zero.

\(\displaystyle \sin \left( n \cdot \frac {2\pi}5 \right) = 0 \iff n \cdot \frac {2\pi}5 = k \cdot \pi \iff n = \frac {5k}2 \)

In order for \(\displaystyle n\) to be an integer, \(\displaystyle 2\) must divide \(\displaystyle k\). Thus \(\displaystyle k = 2p\) where \(\displaystyle p \in \mathbb{Z}\) and consequently \(\displaystyle n = 5p\). With \(\displaystyle 0 \leqslant n \leqslant 4\) we have

\(\displaystyle 0 \leqslant 5p \leqslant 4 \iff 0 \leqslant p \leqslant 0.8 \implies p \in \{ 0 \}\)

So there is one real root and the remaining four are non-real. Similar arguments for \(\displaystyle z^9 = -4\) yield

\(\displaystyle 9v = \pi + n \cdot 2\pi \iff v = \frac {\pi}9 + n \cdot \frac {2\pi}9\)

Equating the imaginary part to zero yields

\(\displaystyle \frac {\pi}9 + n \cdot \frac {2\pi}9 = k \cdot \pi \iff n = \frac {9k-1}2\)

\(\displaystyle 2\) must divide \(\displaystyle 9k-1\) so \(\displaystyle 9k-1 = 2p\). Using \(\displaystyle 0 \leqslant n \leqslant 8\) however yields \(\displaystyle 0 \leqslant p \leqslant 8\). The equation does

**not**have \(\displaystyle 9\) real solutions, so what went wrong?
Last edited: