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Question

Express the following complex in the form r(cos θ + i sin θ):
(i) 1 + i tan α
(ii) tan α − i
(iii) 1 − sin α + i cos α

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Solution

(i) Let z= 1+itan α tan α is periodic with period π. So, let us take α [0,π2)( π2, π]Case I:When α[0,π2)z= 1+itan α z=1+tan2α =sec α 0<α<π2 =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α =tan αβ=α As z lies in the first quadrant . Therefore, arg(z)=β=αThus, z in the polar form is given by z= sec α cosα+isin α Case II:z=1+i tan α z=1+tan2α =sec α π2<α<π =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α =-tan αtan β =tan π-αβ=π-αAs, z lies in the fourth quadrant . arg(z)=-β= α-πThus, z in the polar form is given by z= -sec α cosα-π+isin α-π



(ii) Let z= tan α-i tan α is periodic with period π. So, let us take α [0,π2)( π2, π]Case I:z= tan α-i z=tan2+1 =sec α 0<α<π2 =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α = cot α =cot α =tan π2-αβ=π2-α We can see that Re(z) >0 and Im (z) <0.So, z lies in the fourth quadrant . arg(z)=-β=α- π2Thus, z in the polar form is given by z= sec α cosα-π2+isin α-π2 Case II:z= tan α-i z=tan2+1 =sec α π2<α<π =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α = cot α =-cot α =tan α-π2β=α- π2We can see that Re(z) <0 and Im (z) <0.So, z lies in the third quadrant . arg(z)=π+β= π2+αThus, z in the polar form is given by z= -sec α cosπ2+α+isin π2+α

(iii) Let z= 1-sinα+icosα.sine and cosine functions are periodic functions with period 2π. So, let us take α [0, 2π]Now, z=1-sinα+icosαz=1-sinα2+cos2α=2-sinα=21-sinαz=2cosα2-sinα22=2cosα2-sinα2Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=cosα1-sinα=cos2α2-sin2α2cosα2-sinα22=cosα2+sinα2cosα2-sinα2tanβ=1+tanα21-tanα2=tanπ4+α2Case I: When 0α<π2In this case, we have,cosα2>sinα2 and π4+α2[π4, π2)z=2cosα2-sinα2and tanβ=tanπ4+α2=tanπ4+α2β=π4+α2Clearly, z lies in the first quadrant.Therefore, argz=π4+α2Hence, the polar form of z is 2cosα2-sinα2cosπ4+α2+isinπ4+α2Case II: When π2<α<3π2In this case, we have,cosα2<sinα2 and π4+α2π2, πz=2cosα2-sinα2=-2cosα2-sinα2and tanβ=tanπ4+α2=-tanπ4+α2=tanπ-π4+α2=tan3π4-α2β=3π4-α2Clearly, z lies in the fourth quadrant.Therefore, argz=-β=-3π4-α2=α2-3π4Hence, the polar form of z is -2cosα2-sinα2cosα2-3π4+isinα2-3π4Case III: When 3π2<α<2πIn this case, we have,cosα2<sinα2 and π4+α2π,5π4z=2cosα2-sinα2=-2cosα2-sinα2and tanβ=tanπ4+α2=tanπ4+α2=-tanπ-π4+α2=tanα2-3π4β=α2-3π4Clearly, z lies in the first quadrant.Therefore, argz=β=α2-3π4Hence, the polar form of z is -2cosα2-sinα2cosα2-3π4+isinα2-3π4

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