Calculatrice de nombres complexes

Your input: simplify and calculate different forms of $\left(1 + 3 i\right) \left(5 + i\right)$

Use FOIL to multiply (for steps, see foil calculator), don't forget that $i^2=-1$:

${\color{red}{\left(\left(1 + 3 i\right) \left(5 + i\right)\right)}}={\color{red}{\left(2 + 16 i\right)}}$

Hence, $\left(1 + 3 i\right) \left(5 + i\right)=2 + 16 i$

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 $a=2$ and $b=16$

Thus, $r=\sqrt{\left(2\right)^2+\left(16\right)^2}=2 \sqrt{65}$

Also, $\theta=\operatorname{atan}\left(\frac{16}{2}\right)=\operatorname{atan}{\left(8 \right)}$

Therefore, $2 + 16 i=2 \sqrt{65} \left(\cos{\left(\operatorname{atan}{\left(8 \right)} \right)} + i \sin{\left(\operatorname{atan}{\left(8 \right)} \right)}\right)$

Inverse

The inverse of $2 + 16 i$ is $\frac{1}{2 + 16 i}$

In general case, multiply the expression $\frac{1}{a + i b}$ by the conjugate (the conjugate of $a + i b$ is $a - i b$):

$\frac{1}{a + i b}=\frac{1}{\left(a - i b\right) \left(a + i b\right)} \left(a - i b\right)$

Expand the denominator: $\frac{1}{\left(a - i b\right) \left(a + i b\right)} \left(a - i b\right) = \frac{a - i b}{a^{2} + b^{2}}$

Split:

$\frac{a - i b}{a^{2} + b^{2}}=\frac{a}{a^{2} + b^{2}} - \frac{i b}{a^{2} + b^{2}}$

In our case, $a=2$ and $b=16$

Therefore, ${\color{red}{\left(\frac{1}{2 + 16 i}\right)}}={\color{red}{\left(\frac{1}{130} - \frac{4 i}{65}\right)}}$

Hence, $\frac{1}{2 + 16 i}=\frac{1}{130} - \frac{4 i}{65}$

Conjugate

The conjugate of $a + i b$ is $a - i b$: the conjugate of $2 + 16 i$ is $2 - 16 i$

Modulus

The modulus of $a + i b$ is $\sqrt{a^{2} + b^{2}}$: the modulus of $2 + 16 i$ is $2 \sqrt{65}$

$\left(1 + 3 i\right) \left(5 + i\right)=2 + 16 i=2.0 + 16.0 i$

The polar form of $2 + 16 i$ is $2 \sqrt{65} \left(\cos{\left(\operatorname{atan}{\left(8 \right)} \right)} + i \sin{\left(\operatorname{atan}{\left(8 \right)} \right)}\right)$

The inverse of $2 + 16 i$ is $\frac{1}{2 + 16 i}=\frac{1}{130} - \frac{4 i}{65}\approx 0.00769230769230769 - 0.0615384615384615 i$

The conjugate of $2 + 16 i$ is $2 - 16 i=2.0 - 16.0 i$

The modulus of $2 + 16 i$ is $2 \sqrt{65}\approx 16.1245154965971$