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Solution
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 $$$2 \sqrt{65}$$$
Answer
$$$\left(1 + 3 i\right) \left(5 + i\right)=2 + 16 i=2.0 + 16.0 i$$$
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 $$$2 \sqrt{65}\approx 16.1245154965971$$$