2022-01-17

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

nick1337

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

Vasquez

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 $a-ib$): $\frac{1}{a+ib}=\frac{1}{\left(a-ib\right)\left(a+ib\right)}\left(a-ib\right)$ 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, $a=2$ and $b=3$ Therefore, $\left(\frac{1}{2+3i}\right)=\left(\frac{2}{13}-\frac{3i}{13}\right)$ Hence, $\frac{1}{2+3i}=\frac{2}{13}-\frac{3i}{13}$

alenahelenash

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