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

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Answer 1 · 2018-04-11T09:15:45

$9 [cos (\frac{π}{2}) + i sin (\frac{π}{2})]$ $color(blue)(9[cos(pi/2)+isin(pi/2)]$

For complex numbers given in the form:

$a + b i$ $a+bi$

The polar form will be:

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

Where:

$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$

$θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

$0 + 9 i$ $0+9i$

$r = \sqrt{{(0)}^{2} + {(9)}^{2}} = 9$ $r=sqrt((0)^2+(9)^2)=9$

$θ = arctan (\frac{9}{0})$ $theta=arctan(9/0)$

This is undefined, which tells us that $θ$ $theta$ is either $\frac{π}{2}$ $pi/2$ or $\frac{5 π}{2}$ $(5pi)/2$ .

$9 i$ $9i$ is positive, so this must be in the I and II quadrants. $θ$ $theta$ is therefore $\frac{π}{2}$ $pi/2$

Plugging these values into:

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

$z = 9 [cos (\frac{π}{2}) + i sin (\frac{π}{2})]$ $z=color(blue)(9[cos(pi/2)+isin(pi/2)]$

How do you convert 9i9i to polar form?