Let z be a complex number such that \left | z \right |+z=3+i (where i=\sqrt{-1}). The \left | z \right | is equal to :

Answers (1)
S Sayak

Let z=a+ib

As we know

|z|=\sqrt{a^{2}+b^{2}}

It is given that

\\|z|+z=a+\sqrt{a^{2}+b^{2}}+ib\\ |z|+z=3+i\\ a+\sqrt{a^{2}+b^{2}}+ib=3+i\\

b = 1

\\a+\sqrt{a^{2}+b^{2}}=3\\ a+\sqrt{a^{2}+1}=3\\ a^{2}+1=(3-a)^{2}\\ a^{2}+1=9-6a+a^{2}\\ 6a=8\\ a=\frac{4}{3}

\\|z|=\sqrt{a^{2}+b^{2}}\\ |z|=\sqrt{\left ( \frac{4}{3} \right )^{2}+1^{2}}\\ |z|=\frac{5}{3}

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