$$$15625 + \frac{719413999 i}{1000000000}$$$의 극형식

$$$15625 + \frac{719413999 i}{1000000000}$$$의 극형식을 구하세요.

복소수의 표준형은 $$$15625 + \frac{719413999 i}{1000000000}$$$입니다.

복소수 $$$a + b i$$$에 대해 극형식은 $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$로 주어지며, 여기서 $$$r = \sqrt{a^{2} + b^{2}}$$$ 및 $$$\theta = \operatorname{atan}{\left(\frac{b}{a} \right)}$$$.

다음이 성립한다: $$$a = 15625$$$ 및 $$$b = \frac{719413999}{1000000000}$$$.

따라서, $$$r = \sqrt{15625^{2} + \left(\frac{719413999}{1000000000}\right)^{2}} = \frac{\sqrt{244140625517556501957172001}}{1000000000}.$$$

또한, $$$\theta = \operatorname{atan}{\left(\frac{\frac{719413999}{1000000000}}{15625} \right)} = \operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)}.$$$

따라서 $$$15625 + \frac{719413999 i}{1000000000} = \frac{\sqrt{244140625517556501957172001}}{1000000000} \left(\cos{\left(\operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)} \right)} + i \sin{\left(\operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)} \right)}\right).$$$

$$$15625 + \frac{719413999 i}{1000000000} = \frac{\sqrt{244140625517556501957172001}}{1000000000} \left(\cos{\left(\operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)} \right)} + i \sin{\left(\operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)} \right)}\right) = \frac{\sqrt{244140625517556501957172001}}{1000000000} \left(\cos{\left(\left(\frac{180 \operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)}}{\pi}\right)^{\circ} \right)} + i \sin{\left(\left(\frac{180 \operatorname{atan}{\left(\frac{719413999}{15625000000000} \right)}}{\pi}\right)^{\circ} \right)}\right)$$$A