find the fifth root of 32i that graphs in the first quadrant. ?(cos□° + isin□°)

Explanation:

Step1: Convert (32i) to polar form

A complex number (z = a + bi) in polar form is (z = r(\cos\theta + i\sin\theta)), where (r=\sqrt{a^{2}+b^{2}}) and (\theta=\arctan\left(\frac{b}{a}\right)) (adjusted for the correct quadrant). For (z = 32i), (a = 0), (b = 32). So (r=\sqrt{0^{2}+32^{2}}=32), and since (a = 0) and (b>0), (\theta = 90^{\circ}). Thus, (32i=32\left(\cos90^{\circ}+i\sin90^{\circ}\right)).

Step2: Use De - Moivre's Theorem for roots

De - Moivre's Theorem states that for a complex number (z = r(\cos\theta + i\sin\theta)), the (n)th roots are given by (z_{k}=\sqrt[n]{r}\left(\cos\left(\frac{\theta + 360^{\circ}k}{n}\right)+i\sin\left(\frac{\theta + 360^{\circ}k}{n}\right)\right)), where (k = 0,1,\cdots,n - 1). Here, (n = 5), (r = 32), (\theta=90^{\circ}).

First, find (\sqrt[5]{r}). Since (r = 32=2^{5}), (\sqrt[5]{32}=\sqrt[5]{2^{5}} = 2).

Then, find the angle for the root in the first quadrant. We start with (k = 0) (because for (k = 0), the angle is the smallest positive angle among the roots, and we want the root in the first quadrant). The formula for the angle is (\frac{\theta+360^{\circ}k}{n}). Substituting (n = 5), (\theta = 90^{\circ}), and (k = 0), we get (\frac{90^{\circ}+360^{\circ}\times0}{5}=\frac{90^{\circ}}{5}=18^{\circ}).

Answer:

(2\left(\cos18^{\circ}+i\sin18^{\circ}\right))