How do you convert $9i$ to polar form?

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Answer 1 · 2018-04-11T09:15:45

$color(blue)(9[cos(pi/2)+isin(pi/2)]$

For complex numbers given in the form:

$a+bi$

The polar form will be:

$z=r[cos(theta)+isin(theta)]$

Where:

$r=sqrt(a^2+b^2)$

$theta=arctan(b/a)$

$0+9i$

$r=sqrt((0)^2+(9)^2)=9$

$theta=arctan(9/0)$

This is undefined, which tells us that $theta$ is either $pi/2$ or $(5pi)/2$ .

$9i$ is positive, so this must be in the I and II quadrants. $theta$ is therefore $pi/2$

Plugging these values into:

$z=r[cos(theta)+isin(theta)]$

$z=color(blue)(9[cos(pi/2)+isin(pi/2)]$

How do you convert 9i 9i to polar form?