Question

7) Convert the complex number
z=i-1/cos π/3+ isin π/3 in polar form​

Answer 1

Step-by-step

Z= (i-1)/[cos(π/3) + i.sin(π/3)]

Z= (i-1)/[(1/2) + i.(√3)/2]

Z= (i-1)/[(1+√3.i)/2]

Z= 2(i-1)/[1+√3.i]

Z= [(2i-2)*(1-√3.i)]/[(1+√3.i)*(1- √3.i)]

Z= [2i - 2√3(i^2) - 2 + 2√3.i]/[1 - 3(i^2)]

Z= [(-2 + 2√3) + (2√3 + 2).i]/(1+ 3)

Z= [(-2 + 2√3) + (2√3 + 2).i]/4

Z= (-2 + 2√3)/4 + ((2√3 + 2).i)/4

Z= (-1 + √3)/2 + [(√3 + 1)/2]i

Thus, x=(√3-1)/2, y=(√3+1)/2

|Z| = r = √{[(√3-1)/2]^2 + [(√3-1)/2]^2}

    = √{[(√3-1)^2]/4 + [(√3+1)^2]/4}

    = √{[(3-2√3+1)+(3+2√3+1)]/4}

    = √{[4+4]/4}

    = √{8/4}

    = √2

Now,

 As x > 0 & y > 0

Theta(the symbol) = tan^(-1)[y/x] ...... (tan inverse of y by x)

= tan^(-1){[(√3+1)/2][(√3-1)/2]}

= tan^(-1)[(√3+1)/(√3-1)]

= tan^(-1)[(√3+1)*(√3+1)/(√3-1)*(√3+1)]

= tan^(-1)[(3+2√3+1)/(3-1)]

= tan^(-1)[(4+2√3)/2]

= tan^(-1)[2+√3]

Therefore,

the polar form is,

Z = r[cos(theta) + i sin(theta)]

Z = √2[cos(theta) + i sin(theta)]

where theta = tan^(-1)[2+√3]

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Answer 2

Pls mark it as the brainliest