复数计算器

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 $$$1$$$

$$$i=i=1.0 i$$$

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 $$$1$$$