What is the polar form of #( -1,121 )#?

Question

What is the polar form of #(  -1,121  )#?

Elijah Abram · Accepted Answer

Rectangular form: #(-1,121) hArr # Polar form: #(sqrt(14642),"arccos"(-1/sqrt(14642)))# Radius is given by the Pythagorean Theorem as #color(white)("XXX")r=sqrt(x^2+y^2)# For the given values #(x,y)=(-1,121)# #color(white)("XXX")r=sqrt((-1)^1+121^2) = sqrt(14642)# The given point is in Quadrant II, so it is convenient to use the cos/arccos" functions. #color(white)("XXX")cos(theta)=x/r# #color(white)("XXX")rarr theta = "arccos"(x/r)# #color(white)("XXXXXX")="arccos" ((-1)/sqrt(14642))# Note: If this had any practical applications we would probably convert the above values into (approximate) values (using a calculator) as #color(white)("XXX")(r,theta)~=(121,1.58 " (radians)")#

Paisley Johnson · Answer

undefined The polar form of a complex number $z = a + bi$, where $a$ and $b$ are real numbers, is given by $r(\cos \theta + i\sin \theta)$, where $r = \sqrt{a^2 + b^2}$ is the magnitude of the complex number and $\theta = \arctan\left(\frac{b}{a}\right)$ is the argument of the complex number, measured from the positive x-axis (real axis) towards the y-axis (imaginary axis).

For the complex number corresponding to the point $(-1, 121)$, we interpret $-1$ as the real part $a$ and $121$ as the imaginary part $b$. Therefore, $a = -1$ and $b = 121$.

1. **Magnitude (r):**
$$r = \sqrt{(-1)^2 + (121)^2} = \sqrt{1 + 14641} = \sqrt{14642}$$

2. **Argument ($\theta$):**
$$\theta = \arctan\left(\frac{121}{-1}\right) = \arctan(-121)$$

Because the point is in the second quadrant (where $a$ is negative and $b$ is positive), we need to add $\pi$ radians (or $180^\circ$) to the argument to get the correct angle in standard position (measured counterclockwise from the positive x-axis).

Thus, the argument $\theta$ should be adjusted as $\theta = \arctan(-121) + \pi$. However, this direct calculation might not reflect the intuitive understanding without calculating the actual value of $\arctan(-121)$, which is technically in the fourth quadrant's angle. Since we know it's in the second quadrant, a more correct approach would be acknowledging that the calculator gives us the principal value of the angle, which needs to be interpreted correctly in context. The direct use of $\arctan(-121)$ would typically yield a fourth-quadrant angle due to the negative ratio, thus the need for adjustment by adding $\pi$ radians to ensure it is located correctly in the second quadrant.

Therefore, in terms of a precise numerical calculation:
- For $r$, we have $\sqrt{14642}$.
- For $\theta$, you would compute $\arctan(-121)$ and then adjust by adding $\pi$ to ensure the angle is in the correct quadrant, knowing the value of $\arctan(-121)$ falls in the fourth quadrant by convention, we adjust to correctly place it in the second quadrant.

Finally, the polar form is $r(\cos \theta + i\sin \theta) = \sqrt{14642}(\cos(\arctan(-121) + \pi) + i\sin(\arctan(-121) + \pi))$.

This representation highlights the magnitude and angle, but keep in mind the exact numerical value of $\theta$ requires calculation of $\arctan(-121)$ and adjusting for the quadrant, which typically involves using a calculator or computational tool.