How do you write 2+3i in polar form?

**mariasmiles25** · December 9th 2010, 08:56 AM

**Unknown008** · December 9th 2010, 09:03 AM

The polar form has the general form

$r(\cos\theta + i\sin\theta)$

Where r is the modulus of the complex number and theta the argument of the complex number.

**HallsofIvy** · December 10th 2010, 02:40 AM

The complex number 2+ 3i is represented by the point (2, 3) in the complex plane. The "modulus" unknown008 refers to is the distance from (0, 0) to (2, 3) which is $\sqrt{(2- 0)^2+ (3- 0)^2}$ and the "argument" is the angle the line from (0, 0) to (2, 3) makes with the positive x-axis. That line forms the hypotenuse of a right triangle with "opposite side" the y value, 3, and "near side" the x value, 2. The angle, $\theta$ , is given by $tan(\theta)= \frac{3}{2}$ .

You can derive those relations directly from $2+ 3i= r(cos(\theta)+ i sin(\theta)$ . We must have $rcos(\theta)= 2$ and $rsin(\theta)= 3$ . Square each and add and you get $r^2cos^2(\theta)+ r^2sin^2(\theta)= r^2= 2^2+ 3^2$ . Divide the second equation by the first and you get $\frac{r sin(\theta)}{r cos(\theta)}= \frac{sin(\theta)}{cos(\theta)}= tan(\theta)= \frac{3}{2}$ .

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