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Question

(a) 1+itanα(π<α<π,α±π2)
(b) Find the modulus and argument of the complex number z1=z2z if z=cosϕ+isinϕ.

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Solution

Let rcosθ=1, rsinθ=tanα
Then r=1+tan2α=secα=1|cosθ|
Then cosθ=|cosθ| .....(1)
sinθ=tanα|cosα| .....(2)
b) Substituting z=cosϕ+isinϕ, we get
z1=(cosϕ+isinϕ)2(cosϕ+isinϕ)
=cos2ϕ+isin2ϕcosϕisinϕ
=(cos2ϕcosϕ)+i(sin2ϕsinϕ)
2sin3ϕ2sin(ϕ2)+2icos3ϕ2sinϕ2
2sinϕ2[sin(3ϕ2)+icos3ϕ2]
2sinϕ2[(π2+3ϕ2)+isin(π2+3ϕ2)]
|z|=2sinϕ2
In view of the definition of modulus, we consider three cases:
Case I. Let sinϕ2=0, that is, ϕ=2nπ, n any integer. Than |z1|=0 which implies that z1=0. Thus, for ϕ=2nπ (n, any integer), arg z1 is undefined.
Case II. Let sinϕ2>0 which occurs when
0<ϕ2<π i.e. I, II Quadrant.
or 2nπ<12ϕ<(2n+1)π, that is, when
4nπ<ϕ<(4n+2)π, n any integer. .....(1)
Then |z1|=2sinπ2 and the trigonometrical form of z1 is as
z1=2sinϕ2[cos(π+3ϕ2)+isin(π+3ϕ2)]
Consequently, if ϕ satisfies the condition (1),
then argz1=π+3ϕ2.
Case III. Let sinϕ2<0, that is,
π<ϕ2<2π i.e. III, IV Quadrant.
or (2n+1)π<12ϕ<(2n+2)π
or (4n+2)π<ϕ<(4n+4)π, .....(2)
n any integer.
Then |z1|=2sinϕ2 and the trigonometrical form of z1 is 2sinϕ2[sin3ϕ2icos3ϕ2] by (A)
z1=2sinϕ2[cos3π+3ϕ2+isin3π+3ϕ2]
Hence, when ϕ satisfies (2), we have
argz1=(3π+3ϕ2).

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