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Question

Convert the following in the polar form:
(i) 1+7i(2i)2 (ii) 1+3i12i

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Solution

(i) Here, z=1+7i(2i)2
=1+7i4+i24i
=1+7i414i
=1+7i34i×3+4i3+4i
=3+4i+21i+28i232+42
=3+4i+21i2832+42=25+25i25

=1+i
Let rcosθ=1 and rsinθ=1

On squaring and adding, we obtain

r2(cos2θ+sin2θ)=1+1

r2(cos2θ+sin2θ)=2

r2=2 (cos2θ+sin2θ=1)

r=2 (As r>0 )

2cosθ=1 and 2sinθ=1

cosθ=12 and

sinθ=12

θ=ππ4=3π4

(As θ lies in II quadrant )

z=rcosθ+irsinθ

=2cos3π4+i2sin3π4=2(cos3π4+isin3π4)

This is the required polar form.
(ii) Here,
z=1+3i12i
=1+3i12i×1+2i1+2i
=1+2i+3i61+4
=5+5i5

=1+i

Let rcosθ=1 and rsinθ=1

On squaring and adding, we obtain

r2(cos2θ+sin2θ)=1+1

r2(cos2θ+sin2θ)=2

r2=2 (cos2θ+sin2θ=1)

r=2 (As r>0 )

2cosθ=1 and 2sinθ=1

cosθ=12 and

sinθ=12

θ=ππ4=3π4

(As θ lies in II quadrant )

z=rcosθ+irsinθ

=2cos3π4+i2sin3π4=2(cos3π4+isin3π4)

This is the required polar form.

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