Equation with complex numbers - number of solutions

Question

Find out how many solutions in quadrant 3 of the trigonometric circle has the equation $z^{100}=1+i$.

I found that there are $25$ solutions in quadrant 3. My solution is to find $r \geq0$ where $r$ is natural number so $2r+1\le\frac{8k+1}{400}\le\frac{4r+3}{2}$ ($k$ takes the values from $0$ to $99$). From $k=75$ to $99$. I find that exist $r$.

$z^{100}=\sqrt2(\cos(\pi/4)+i \sin(\pi/4))$

Is it ok?

Answer 1

Note that

• $\arg(1+i)=\frac{\pi}4$

thus

• $\arg(z_k)=\frac{\pi}{400}+\frac{2\pi k}{100}$ with $k=0$ to $99$.

thus we need to consider

$$\pi\le \frac{\pi+8k\pi}{400} \le \frac {3\pi}{2}\iff 1\le \frac{1+8k}{400} \le \frac {3}{2}\iff399\le8k\le599 \iff k\in [50,74]$$

Answer 2

Hint

$$z^{100}=\sqrt 2 e^{i\left(\frac {\pi}{4}+2k\pi\right)}$$ Hence $$z=\sqrt[200] {2} e^{i\left(\frac {\pi}{400}+\frac {k\pi}{50}\right)}$$ Here $k$ goes from $0$ to $99$. We need to find k when the argument falls in third quadrant. Once you write down a few of them you will notice a pattern just give a try

Or else solve $$\pi\le \frac {\pi}{400}+\frac {k\pi}{50}\le \frac {3\pi}{2}$$

Giving $$399\le 8k\le 599$$to find appropriate integer $k$

Hence $k\in [50,74]$