How do you convert #8i# to polar form?

Question

Victoria Adebayo · Accepted Answer

#8i=8(cos(pi/2)+isin(pi/2))# #a+bi# in polar form is #rcostheta+irsintheta#, where #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)# #8i=0+8i# hence #r=sqrt(0^2+8^2)=8# and #theta=arctan(8/0)=arctanoo# and one can say #theta=pi/2# #8i=8(cos(pi/2)+isin(pi/2))#

Brooklyn Anderson · Answer

undefined To convert $ 8i $ to polar form, we need to find its magnitude (r) and argument (θ).

First, the magnitude (r) can be found using the formula:
$$ r = \sqrt{a^2 + b^2} $$
where $ a $ is the real part and $ b $ is the imaginary part. For $ 8i $, the real part is 0 and the imaginary part is 8, so:
$$ r = \sqrt{0^2 + 8^2} = \sqrt{64} = 8 $$

Second, the argument (θ) can be found using the formula:
$$ \theta = \arctan\left(\frac{b}{a}\right) $$
where $ a $ is the real part and $ b $ is the imaginary part. For $ 8i $, the real part is 0 and the imaginary part is 8, so:
$$ \theta = \arctan\left(\frac{8}{0}\right) $$

Since the real part is 0, we need to be careful when calculating the argument. In this case, $ \theta $ is either $ \frac{\pi}{2} $ or $ -\frac{\pi}{2} $, depending on the quadrant in which the complex number lies.

Therefore, in polar form, $ 8i $ can be expressed as:
$$ 8i = 8(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) $$
or
$$ 8i = 8(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})) $$