How is -1+i√3 written in polar form?

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Answer 1 · 2018-04-26T12:30:47

See explanation.

First we need to calculate the module:

#|z|=sqrt(a^2+b^2)=sqrt((-1)^2+sqrt(3)^2)=sqrt(1+3)=2#

Now we have:

Here #|z|=2#, so

The angle, for which #cosvarphi=-1/2# is #varphi=120^o#, so now we can write the complex number in trigonometric form:

#z=2*(cos120^o + isin120^o)#

Answer 2 · 2018-04-26T12:42:22

In polar form expressed as #2(cos 2.0944+i sin 2.0944)#

Let #Z=a+i b ; Z=-1+ i sqrt 3 ; a= -1 ,b =sqrt 3 # ;

#Z=-1+ i sqrt 3 # is in 2nd quadrant

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt((-1)^2+ (sqrt3)^2)) =2 #

#tan alpha =|b/a|= (sqrt 3/ -1) = sqrt 3 :. alpha =tan^-1(sqrt3)#

or # alpha~~ 1.047198; theta# is on #2# nd quadrant

# :. theta=pi- alpha= 2.094395 :.# Argument : # theta ~~2.0944 #

In polar form expressed as

# |Z|*(cos theta+i sin theta) or 2(cos 2.0944+i sin 2.0944)#[Ans]