# Complex Number Calculator

The calculator will try to simplify any complex expression, with steps shown. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus, and inverse of the complex number.

Enter an expression:

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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 $2 \sqrt{65}$

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