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Komplexzahlenrechner

Operationen mit komplexen Zahlen Schritt für Schritt durchführen

Der Rechner versucht, jeden beliebigen komplexen Ausdruck zu vereinfachen, mit angezeigten Lösungsschritten. Er führt Addition, Subtraktion, Multiplikation, Division und Potenzieren durch und bestimmt außerdem die Polarform, das Konjugierte, den Betrag und den Kehrwert der komplexen Zahl.

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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 $$$2 \sqrt{65}\approx 16.1245154965971$$$