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Solve z+|z|=2+8i to Find |z| | Complex Numbers JEE

JEE Maths question with a full step-by-step solution.

Question
If the complex number zz satisfies z+∣z∣=2+8iz + |z| = 2 + 8i, then the value of ∣z∣|z| will be
A88
B1717correct
C1515
D2424
Solution
Step 1: Write z=a+ibz = a + ib, so ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2} (a real number). The equation becomes:
a+ib+a2+b2=2+8ia + ib + \sqrt{a^2 + b^2} = 2 + 8i
 Step 2: Separate into real and imaginary parts. The imaginary part of the left side is just bb (the modulus is real), so:
Imaginary: b=8\text{Imaginary: } b = 8
Real: a+a2+b2=2\text{Real: } a + \sqrt{a^2 + b^2} = 2
 Step 3: Substitute b=8b = 8 into the real equation:
a+a2+64=2  ⟹  a2+64=2−aa + \sqrt{a^2 + 64} = 2 \implies \sqrt{a^2 + 64} = 2 - a
 Step 4: Square both sides (valid since the left side is non-negative, which also requires 2−a≥02 - a \geq 0):
a2+64=(2−a)2=4−4a+a2a^2 + 64 = (2 - a)^2 = 4 - 4a + a^2
64=4−4a  ⟹  4a=−60  ⟹  a=−1564 = 4 - 4a \implies 4a = -60 \implies a = -15
 (The requirement 2−a≥02 - a \geq 0 holds since 2−(−15)=17>02 - (-15) = 17 > 0.) Step 5: Compute the modulus:
∣z∣=a2+b2=(−15)2+82=225+64=289=17|z| = \sqrt{a^2 + b^2} = \sqrt{(-15)^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17
 Correct answer: (2)
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