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Question

Question: How do you convert \(9i\) to polar form?...

How do you convert 9i9i to polar form?

Explanation

Solution

We will use the polar representation of a complex number to solve this question. The polar representation of complex number a+iba+ib is given as z=r(cosθ+isinθ)z=r\left( \cos \theta +i\sin \theta \right), where r is the distance between the point from origin and is expressed as r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}} and θ\theta is the angle which is expressed as θ=tan1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) .

Complete step by step answer:
We have been given a complex number 9i9i.
We have to convert it into the polar form.
Now, we know that the polar form of a complex number a+iba+ib is given as z=r(cosθ+isinθ)z=r\left( \cos \theta +i\sin \theta \right), where r is the distance between the point from origin and is expressed as r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}} and θ\theta is the angle which is expressed as θ=tan1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) .
Now, we have a complex number 9i9i, we can write it in standard form as
0+9i\Rightarrow 0+9i
Now, we get a=0,b=9a=0,b=9
Now, let us find the value of r, substituting the values in the formula we will get
r=02+92\Rightarrow r=\sqrt{{{0}^{2}}+{{9}^{2}}}
Now, simplifying the above obtained equation we will get
r=92 r=9 \begin{aligned} & \Rightarrow r=\sqrt{{{9}^{2}}} \\\ & \Rightarrow r=9 \\\ \end{aligned}
Now, θ=tan1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)
Substituting the values we will get
θ=tan1(90)\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{9}{0} \right)
Now, simplifying the above obtained expression we will get
θ=tan1() θ=π2 \begin{aligned} & \Rightarrow \theta ={{\tan }^{-1}}\left( \infty \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{2} \\\ \end{aligned}
Now, the polar form of the given complex number will be
z=9(cosπ2+isinπ2)z=9\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)

Hence above is the required polar form of 9i9i.

Note: In this particular question the point to be noted is that tangent function has undefined or infinite value on π2\dfrac{\pi }{2} and 5π2\dfrac{5\pi }{2}. But the given value 9i9i is positive it means θ\theta must have positive value and lies in the first and second quadrant so we will take the value of θ\theta as π2\dfrac{\pi }{2}.