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

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- Dec 9th 2010, 08:56 AMmariasmiles25How do you write 2+3i in polar form?
How do you write 2+3i in polar form?

- Dec 9th 2010, 09:03 AMUnknown008
The polar form has the general form

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

Where r is the modulus of the complex number and theta the argument of the complex number. (Smile) - Dec 10th 2010, 02:40 AMHallsofIvy
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 $\displaystyle \sqrt{(2- 0)^2+ (3- 0)^2}$ 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, $\displaystyle \theta$, is given by $\displaystyle tan(\theta)= \frac{3}{2}$.

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