Question

(iv) If u + iv= 2+i /z+3 and z=x+iy then find u, v​

Answer 1

Answer:

z=e

iα

and w=e

iβ

z

2

+w

2

=1⇒e

i2α

+e

i2β

=1+i0

⇒cos2α+cos2β=1 and sin2α+sin2β=0

⇒2cos(α+β)cos(α−β)=1 .............eq(i)

and 2sin(α+β)cos(α−β)=0 ................eq(ii)

Now Dividing eq(ii) by eq(i)

⇒tan(α+β)=0⇒α+β=nπ

for α+β=0 ,

we have cos2α=

2

1

⇒ 4 pairs (α,β) for α∈[0,2π) [Note: 2 pairs (α,β) + 2 pairs (β,α)= 4 pairs]

for α+β=π ,

we have cos2α=

2

1

⇒ 4 pairs (α,β) for α∈[0,2π) [Note: 2 pairs (α,β) + 2 pairs (β,α)= 4 pairs]

∴ Total number of ordered pairs (z,w) is 8 pairs

Step-by-step explanation:

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