복소수 계산기
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Solution
Your input: simplify and calculate different forms of $$$i$$$
The expression is already simplified.
Polar form
For a complex number $$$a+bi$$$, polar form is given by $$$r(\cos(\theta)+i \sin(\theta))$$$, where $$$r=\sqrt{a^2+b^2}$$$ and $$$\theta=\operatorname{atan}\left(\frac{b}{a}\right)$$$
We have that $$$a=0$$$ and $$$b=1$$$
Thus, $$$r=\sqrt{\left(0\right)^2+\left(1\right)^2}=1$$$
Also, $$$\theta=\operatorname{atan}\left(\frac{1}{0}\right)=\frac{\pi}{2}$$$
Therefore, $$$i=\cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)}$$$
Inverse
The inverse of $$$i$$$ is $$$\frac{1}{i}$$$
Multiply and divide by $$$i$$$ (keep in mind that $$$i^2=-1$$$):
$$${\color{red}{\left(\frac{1}{i}\right)}}={\color{red}{\left(- i\right)}}$$$
Hence, $$$\frac{1}{i}=- i$$$
Conjugate
The conjugate of $$$a + i b$$$ is $$$a - i b$$$: the conjugate of $$$i$$$ is $$$- i$$$
Modulus
The modulus of $$$a + i b$$$ is $$$\sqrt{a^{2} + b^{2}}$$$: the modulus of $$$i$$$ is $$$1$$$
Answer
$$$i=i=1.0 i$$$
The polar form of $$$i$$$ is $$$\cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)}$$$
The inverse of $$$i$$$ is $$$\frac{1}{i}=- i=- 1.0 i$$$
The conjugate of $$$i$$$ is $$$- i=- 1.0 i$$$
The modulus of $$$i$$$ is $$$1$$$