If |z + 1| = αz + β (i + 1) and
– 2i, find α + β.
If |z + 1| = αz + β (i + 1) and – 2i, find α + β.
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1 Answer
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.(1)
–2α + β = 0 …(2)
Solving (1) and (2)
a = 1
b = 2
-> a + b = 3
Similar Questions for you
|z| = 0 (not acceptable)
|z| = 1
|z|2 = 1
Given : x2 – 70x + l = 0
->Let roots be a and b
->b = 70 – a
->= a (70 – a)
l is not divisible by 2 and 3
->a = 5, b = 65
->
z1 + z2 = 5
⇒ 20 + 15i = 125 – 15z1z2
⇒ 3z1z2 = 25 – 4 – 3i
3z1z2 = 21– 3i
z1⋅z2 = 7 – i
(z1 + z2)2 = 25
= 11 + 2i
= 121 − 4 + 44i
⇒
⇒ = 117 + 44i − 2(49 −1−14i )
= 21 + 72i
⇒
a = 1 > 0 and D < 0
4 (3k – 1)2 – 4 (8k2 – 7) < 0
K = 3
Let z be those complex numbers which satisfy
If the maximum value of then the value of (a + b) is…….
->Represent a circle
->Represent a line X – y
So max |z + 1|2 = AQ2
Hence a + b = 48

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