Question

find the graph of this function as the value of n increases, starting from n = 1. f(n) = (3/5 + 4/5 i)^n remember |a + bi| = √(a² + b²)

Explanation:

Step1: Find the modulus of the complex number

The complex number is \( z = \frac{3}{5}+\frac{4}{5}i \). Using the formula \( |a + bi|=\sqrt{a^{2}+b^{2}} \), we have \( |z|=\sqrt{(\frac{3}{5})^{2}+(\frac{4}{5})^{2}}=\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{\frac{25}{25}} = 1 \). So, \( |z^{n}|=|z|^{n}=1^{n}=1 \) for all \( n \), which means all points \( f(n) \) lie on the unit circle.

Step2: Find the argument of the complex number

The argument \( \theta \) of \( z=\frac{3}{5}+\frac{4}{5}i \) satisfies \( \tan\theta=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3} \), so \( \theta=\arctan(\frac{4}{3}) \), which is an acute angle in the first quadrant. When we raise \( z \) to the power \( n \), the argument of \( z^{n} \) is \( n\theta \) (by De - Moivre's theorem: \( (r(\cos\theta + i\sin\theta))^{n}=r^{n}(\cos(n\theta)+i\sin(n\theta)) \), here \( r = 1 \)). As \( n \) increases, the argument \( n\theta \) increases, and the points \( z^{n} \) are equally - spaced (since the angle between consecutive points \( z^{n + 1}\) and \( z^{n}\) is \( \theta \)) on the unit circle.
Now, let's analyze the three graphs:

• The first graph: The points seem to be symmetric in a way that does not match the equal - spacing with an acute - angled argument in the first quadrant.
• The second graph: The points are equally spaced on the unit circle, and the starting angle (for \( n = 1\)) is in the first quadrant, which matches our calculation of the argument of \( z=\frac{3}{5}+\frac{4}{5}i \) (acute angle in the first quadrant) and the fact that the modulus is 1.
• The third graph: The spacing or the starting position of the points does not match the properties of \( z^{n}\) with \( z=\frac{3}{5}+\frac{4}{5}i \).

Answer:

The Middle Graph