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Calculatrice de nombres complexes

Effectuez des opérations sur les nombres complexes étape par étape

La calculatrice essaiera de simplifier toute expression complexe, en affichant les étapes. Elle effectuera l'addition, la soustraction, la multiplication, la division, l'élévation à une puissance, et déterminera également la forme polaire, le conjugué, le module et l'inverse du nombre complexe.

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