Since, ∣z∣+z=3+i Let z=a+ib, then ∣z∣+z=3+i⇒a2+b2+a+ib=3+i Compare real and imaginary coefficients on both sides b=1,a2+b2+a=3a2+1=3−aa2+1=a2+9−6a6a=8⇒a=34 Then, ∣z∣=(34)2+1=916+1=35
Let z be a complex number such that ∣z∣+z=3+i ( where i=−1) Then ∣z∣ is equal to :
Held on 11 Jan 2019 · Verified 6 Jul 2026.
334
35
441
45
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