Question

find the polar form of. 1÷1+i and 1+7i÷(2-i)square​

Answer 1

Answer:

1.

$\frac{1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 - i}{ {1}^{2} - {i}^{2} } = \frac{1 - i}{2} \\ \\ \frac{1}{ \sqrt{2} } ( \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } i) \\ \\ = \frac{1}{ \sqrt{2} } ( \cos( \frac{7\pi}{4} ) + i \sin( \frac{7\pi}{4} ) )$

2.

$\frac{1 + 7i}{(2 - i) {}^{2} } \\ \\ = \frac{1 + 7i}{ {2}^{2} + {i}^{2} - 4i } \\ \\ = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} \\ \\ = \frac{3 + 4i + 21i - 28}{9 + 16} \\ \\ = \frac{25 + 25i}{25} \\ \\ = 1 + i \\ \\ = \sqrt{2} ( \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } i) \\ \\ = \sqrt{2} ( \cos( \frac{\pi}{4} ) + i \sin( \frac{\pi}{4} ) )$