2022-01-17

How do you simplify $|2+3i|?$

nick1337

Expert2022-01-18Added 573 answers

Step 1
Given: $|2+3i|$
$|a+bi|=\sqrt{{a}^{2}+{b}^{2}}$
So,
$|2+3i|=\sqrt{{2}^{2}+{3}^{2}}$
$\Rightarrow \sqrt{4+9}$
$\Rightarrow \sqrt{13}$
Hence, $|2+3i|=\sqrt{13}$

Vasquez

Skilled2022-01-18Added 457 answers

Step 1
The inverse of $2+3i$ is $\frac{1}{2+3i}$
In general case, multiply the expression $\frac{1}{a+bi}$ 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, $a=2$ and $b=3$
Therefore, $(\frac{1}{2+3i})=(\frac{2}{13}-\frac{3i}{13})$
Hence, $\frac{1}{2+3i}=\frac{2}{13}-\frac{3i}{13}$

alenahelenash

Skilled2022-01-24Added 366 answers

Step 1
We have that $a=2$ and $b=3$
Thus,
$r=\sqrt{(2{)}^{2}+(3{)}^{2}}=\sqrt{13}$
Also,
$\theta =a\mathrm{tan}(\frac{3}{2})=a\mathrm{tan}(\frac{3}{2})$
Therefore,
$2+3i=\sqrt{13}(\mathrm{cos}(a\mathrm{tan}(\frac{3}{2}))+i\mathrm{sin}(a\mathrm{tan}(\frac{3}{2})))$