Given: ∣z+1∣=αz+β(1+i),z=21−2i
⇒∣21−2i+1∣=α(21−2i)+β(1+i)
⇒∣23−2i∣=2α−2αi+β+βi
⇒49+4=(2α+β)+i(−2α+β)
Now, on comparing real and imaginary part we get,
⇒25=(2α+β)+i(−2α+β)
⇒2α=β,25=(2α+β)
⇒2α=β,25=25α
⇒α=1,β=2
⇒α+β=3
If z=21−2i, is such that ∣z+1∣=αz+β(1+i),i=−1 and α,β∈R, then α+β is equal to
Held on 29 Jan 2024 · Verified 6 Jul 2026.
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