No AI is used
  1. Hem
  2. Kalkylatorer
  3. Kalkylatorer: Algebra II
  4. Precalculus-kalkylator

Kalkylator för komplexa tal

Utför operationer med komplexa tal steg för steg

Kalkylatorn försöker förenkla vilket komplext uttryck som helst och visar stegen. Den utför addition, subtraktion, multiplikation, division och upphöjning till potens samt bestämmer den polära formen, komplexkonjugatet, modulen och inversen för det komplexa talet.

Enter an expression:

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

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