Questions: Find the square root of -9 i that graphs in the second quadrant. 3(cos [?]°+i sin [?]°)

Solution Steps

To find the square root of a complex number, we can use the polar form of the complex number. The given complex number is \(-9i\). We need to convert this to polar form, find the square roots, and then determine which one lies in the second quadrant.

1. Convert \(-9i\) to polar form.
2. Use the formula for the square root of a complex number in polar form.
3. Identify the root that lies in the second quadrant.

Given the complex number \( -9i \), we first convert it to polar form. The magnitude \( r \) and the angle \( \theta \) are calculated as follows: \[ r = | -9i | = 9 \] \[ \theta = \arg(-9i) = -\frac{\pi}{2} \]

To find the square roots, we use the polar form of the complex number. The square roots are given by: \[ \sqrt{r} = \sqrt{9} = 3 \] \[ \theta_1 = \frac{\theta}{2} = \frac{-\frac{\pi}{2}}{2} = -\frac{\pi}{4} \] \[ \theta_2 = \frac{\theta}{2} + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} \]

We convert the polar coordinates back to rectangular form: \[ \text{root}_1 = 3 \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) = 3 \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = 2.1213 - 2.1213i \] \[ \text{root}_2 = 3 \left( \cos\left(\frac{3\pi}{4}\right) + i \sin\left(\frac{3\pi}{4}\right) \right) = 3 \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -2.1213 + 2.1213i \]

The second quadrant is defined by \( x < 0 \) and \( y > 0 \). Among the roots, \(\text{root}_2\) satisfies this condition: \[ \text{root}_2 = -2.1213 + 2.1213i \]

The magnitude and angle of \(\text{root}_2\) are: \[ \text{magnitude} = 3 \] \[ \text{angle} = \frac{3\pi}{4} \times \frac{180}{\pi} = 135^\circ \]

Final Answer