2022-01-17

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

nick1337

Expert2022-01-18Added 777 answers

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

Expert2022-01-18Added 669 answers

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

alenahelenash

Expert2022-01-24Added 556 answers

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 $a-ib$ ):
$\frac{1}{a+ib}=\frac{1}{(a-ib)(a+ib)}(a-ib)$
Expand the denominator:
$\frac{1}{(a-ib)(a+ib)}(a-ib)=\frac{a-ib}{{a}^{2}+{b}^{2}}$
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,
$(\frac{1}{1+3i})=(\frac{1}{10}-\frac{3i|}{10})$
Hence, $\frac{1}{1+3i}=\frac{1}{10}-\frac{3i}{10}$

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