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Question
Solve the equation |z| = z + 1 + 2i.
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Solution
Given that: |z| = z + 1 + 2i
Let z = x + iy
|z| = (z + 1) + 2i
Squaring both sides
|z|2 = |z + 1|2 + 4i2 + 4(z + 1)i
⇒ |z|2 = |z|2 + 1 + 2z – 4 + 4(z + 1)i
⇒ 0 = –3 + 2z + 4(z + 1)i
⇒ 3 – 2z – 4(z + 1)i = 0
⇒ 3 – 2(x + yi) – 4[x + yi + 1]i = 0
⇒ 3 – 2x – 2yi – 4xi – 4yi2 – 4i = 0
⇒ 3 – 2x + 4y – 2yi – 4i – 4xi = 0
⇒ (3 – 2x + 4y) – i(2y + 4x + 4) = 0
⇒ 3 – 2x + 4y = 0
⇒ 2x – 4y = 3 .....(i)
And 4x + 2y + 4 = 0
⇒ 2x + y = –2 .....(ii)
Solving equation (i) and (ii), we get
y = –1 and x = `-1/2`
Hence, the value of z = x + yi = `(- 1/2 - i)`.
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