Question - Complex Number Conversion to Polar Form

Let's simplify \(\frac{1+7i}{(2-i)^2}\):

\(= \frac{1+7i}{(2-i)(2-i)}\)

\(= \frac{1+7i}{4 - 2i - 2i + i^2}\)

\(= \frac{1+7i}{4 - 4i - 1}\) (since \(i^2 = -1\))

\(= \frac{1+7i}{3 - 4i}\)

Multiply the numerator and the denominator by the conjugate of the denominator:

\(= \frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}\)

\(= \frac{3 + 4i + 21i + 28i^2}{9 + 12i - 12i - 16i^2}\)

\(= \frac{3 + 25i - 28}{9 + 16}\) (since \(i^2 = -1\))

\(= \frac{-25 + 25i}{25}\)

\(= -1 + i\)

The magnitude \(r\) is given by \(r = \sqrt{(-1)^2 + (1)^2}\):

\(= \sqrt{1 + 1}\)

\(= \sqrt{2}\)

The angle \(\theta\) can be found from \(\tan(\theta) = \frac{1}{-1}\):

\(\theta = \arctan(-1)\)

We notice that the complex number lies in the second quadrant, hence \(\theta = \pi + \arctan(-1)\)

\(\theta = \pi - \frac{\pi}{4}\)

\(\theta = \frac{3\pi}{4}\)

The polar form of the complex number is \(r(\cos(\theta) + i\sin(\theta))\):

\(= \sqrt{2} \left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right)\)