Rekenmachine voor complexe getallen
Voer stap voor stap bewerkingen uit op complexe getallen
De rekenmachine zal proberen elke complexe uitdrukking te vereenvoudigen, waarbij de stappen worden getoond. De rekenmachine zal optelling, aftrekking, vermenigvuldiging, deling en machtsverheffing uitvoeren, en zal ook de poolvorm, het complex geconjugeerde, de modulus en de reciproke van het complexe getal bepalen.
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 $$$2 \sqrt{65}\approx 16.1245154965971$$$