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

1 Answer

Shwetank Mauria

Apr 13, 2016

#8i=8(cos(pi/2)+isin(pi/2))#

Explanation:

#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))#

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