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z= i-1/(1/2+√3/2)
z= 2(i-1)/1+√3i
z= 2(i-1)(1-√3i)/(1+√3i)(1-√3i)
z= (√3-1)/2 + (√3+1)/2
in polar form z=√2(cos 5π/12 + isin 5π/12)
PLEASE SELECT MY ANSWERS AS BRAINLIEST
z= 2(i-1)/1+√3i
z= 2(i-1)(1-√3i)/(1+√3i)(1-√3i)
z= (√3-1)/2 + (√3+1)/2
in polar form z=√2(cos 5π/12 + isin 5π/12)
PLEASE SELECT MY ANSWERS AS BRAINLIEST
Answered by
1
we have ,z=i-1/1/2+√3/2i
=2(i-1)/1+√3i*1-√3i/1-√3i
=2(i+√3-1+√3i)/1+3
=√3-1/2+√3+1/2i
now we get √3-1/2=rcostheta ,√3+1/2 rsin theta
by squaring and adding we get
r^2=[√3-1/2]^2+[√3+1/2]^2=2[(√3)^2+1)/4=2*4/4=2
hence r=√2which gives cos theta =√3-1/2√2,sin theta =√3+1/2√2
now the polar form is
√2(cos 5Π/12+i sin 5Π/12)
=2(i-1)/1+√3i*1-√3i/1-√3i
=2(i+√3-1+√3i)/1+3
=√3-1/2+√3+1/2i
now we get √3-1/2=rcostheta ,√3+1/2 rsin theta
by squaring and adding we get
r^2=[√3-1/2]^2+[√3+1/2]^2=2[(√3)^2+1)/4=2*4/4=2
hence r=√2which gives cos theta =√3-1/2√2,sin theta =√3+1/2√2
now the polar form is
√2(cos 5Π/12+i sin 5Π/12)
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