2022-01-17

How do you find the absolute value of $1+3i$?

nick1337

Step 1 $|1+3i|=\sqrt{10}$ The absolute value of a complex number is its distance from the origin 0 in the complex plane. By Pythagoras

Vasquez

Step 1 For a complex number $a+bi$, polar form is given by $r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)$, where $r=\sqrt{{a}^{2}+{b}^{2}}$ and $\theta =a\mathrm{tan}\left(\frac{b}{a}\right)$ We have that $a=1$ and $b=3$ Thus, $r=\sqrt{\left(1{\right)}^{2}+\left(3{\right)}^{2}}=\sqrt{10}$ Also, $\theta =a\mathrm{tan}\left(\frac{3}{1}\right)=a\mathrm{tan}\left(3\right)$ Therefore, $1+3i=\sqrt{10}\left(\mathrm{cos}\left(a\mathrm{tan}\left(3\right)\right)+i\mathrm{sin}\left(a\mathrm{tan}\left(3\right)\right)\right)$

alenahelenash

Step 1 The inverse of $1+3i$ is $\frac{1}{1+3i}$ In general case, multiply the expression $\frac{1}{a+ib}$ by the conjugate (the conjugate of $a+ib$ is $a-ib$): $\frac{1}{a+ib}=\frac{1}{\left(a-ib\right)\left(a+ib\right)}\left(a-ib\right)$ Expand the denominator: $\frac{1}{\left(a-ib\right)\left(a+ib\right)}\left(a-ib\right)=\frac{a-ib}{{a}^{2}+{b}^{2}}$ Split: $\frac{a-ib}{{a}^{2}+{b}^{2}}=\frac{a}{{a}^{2}+{b}^{2}}-\frac{ib}{{a}^{2}+{b}^{2}}$ In our case, $a=1$ and $b=3$ Therefore, $\left(\frac{1}{1+3i}\right)=\left(\frac{1}{10}-\frac{3i|}{10}\right)$ Hence, $\frac{1}{1+3i}=\frac{1}{10}-\frac{3i}{10}$