How can I write $2+5i$ in polar form?

Question

Original problem:

Write $2+5i$ in polar form.

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My attempt:

$|2+5i|=\sqrt{4+25}=\sqrt{29}$

$\arg(2+5i)=\arctan{\frac{5}{2}}=\theta$

$2+5i=\sqrt{29}(\cos(\theta)+i\sin(\theta))$

But how do I calculate $\arctan({\frac{5}{2}})$ in radians?

I know $\tan(\theta)=\frac{5}{2}$ and this implies that if we have a rectangle triangle $ABC$ then $\overline{AB}=5$ and $\overline{BC}=2$.

Here I'm stuck here.

Answer 1

As mentioned in comments above, you do not need to find the value of $\theta$, namely, you don't need to find the value of $\arctan \frac{5}{2}$. Instead, what you need here are the following two quantities: $$ \cos\theta,\quad \sin\theta. $$ In your right triangle, $$ \sin\theta=\frac{5}{\overline{AC}},\quad \cos\theta=\frac{2}{\overline{AC}}. $$ But you could find the value of $\overline{AC}$ using the Pythagorean theorem.

Answer 2

What we need is the values of $\cos\arctan(5/2)$ and $\sin\arctan(5/2)$, and not $\arctan(5/2)$ itself. This can be done easily in general: draw a right angled triangle with hypotenuse $\sqrt{1+x^2}$ and one of the other sides of length $x$, then the other side has length $1$ and it is easy to identify the angle whose tangent is $x$. So the $\sin,\cos$ of that angle can be found as a simple exercise in trigonometric ratios. In general it is easy to see that $$\sin\arctan(x)=\frac{x}{\sqrt{x^2+1}},\cos\arctan(x)=\frac1{\sqrt{x^2+1}}.$$

Answer 3

USE: $z = re^{i\theta}$

$2+5i$ = ?

Using modulus which is $r = \sqrt{x^2+y^2}$

$r = \sqrt{(2)^2+(5)^2}$ = $\sqrt{29}$

$\theta = \arctan({\frac{5}{2}})$

$2+5i = \sqrt{29}e^{i\arctan({\frac{5}{2}})}$