# What is the trigonometric form of  (1+5i) ?

Jan 20, 2016

$1 + 5 i = \sqrt{26} \left\{\cos \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right\}$

#### Explanation:

Trigonometric form of any complex number of the form $a + i b$ is given by r(cos(theta)+isin(theta). If one is able to find $r$ and $\theta$ then the job is done.

Let us see how we go about find $r$ and $\theta$ let me teach you

Say our complex number is $z = a + i b$

We want to represent it as $r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$ let us distribute the $r$ first.

$a + i b = r \cos \left(\theta\right) + i \sin \left(\theta\right)$

Let us equate the real parts

$a = r \cos \left(\theta\right)$
Squaring both sides.
${a}^{2} = {r}^{2} {\cos}^{2} \left(\theta\right)$

Equating the imaginary parts we get
$b = r \sin \left(\theta\right)$
Squaring both sides.
${b}^{2} = {r}^{2} {\sin}^{2} \left(\theta\right)$

Adding ${a}^{2}$ and ${b}^{2}$
${a}^{2} + {b}^{2} = {r}^{2} {\cos}^{2} \left(\theta\right) + {r}^{2} {\sin}^{2} \left(\theta\right)$
${a}^{2} + {b}^{2} = {r}^{2} \left({\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right)$ quadcolor(green)("factored out GCF " r^2
${a}^{2} + {b}^{2} = {r}^{2} \left(1\right)$quadcolor(green) (quad cos^2(theta)+sin^2(theta)=1

${a}^{2} + {b}^{2} = {r}^{2}$

$\implies r = \sqrt{{a}^{2} + {b}^{2}}$ $\quad \textcolor{g r e e n}{r}$ $\textcolor{g r e e n}{\text{ is a distance, so, it would be positive}}$

Dividing$\frac{b}{a}$ we get

$\frac{b}{a} = \frac{r \sin \left(\theta\right)}{r \cos \left(\theta\right)}$
$\frac{b}{a} = \tan \left(\theta\right)$

$\implies {\tan}^{-} 1 \left(\frac{b}{a}\right) = \theta$

We get $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

Now a question would arise is it needed to do so many steps. The answer to that it's not required .

We just need to know

$r = \sqrt{{a}^{2} + {b}^{2}} \quad \theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

Now let us get back to our problem.

$1 + 5 i$

$r = \sqrt{{1}^{2} + {5}^{2}}$
$r = \sqrt{1 + 25}$
$r = \sqrt{26}$

$\theta = {\tan}^{-} 1 \left(\frac{5}{1}\right)$
$\theta = {\tan}^{-} 1 \left(5\right)$

The trigonometric form
$1 + 5 i = \sqrt{26} \left\{\cos \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right\}$