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Home » Solve that equation |z| = z + 1 + 2i.
Solve that equation |z| = z + 1 + 2i.
CBSE | Class 11 | Complex Numbers and Quadratic Equations | Maths | NCERT Exemplar
Solve that equation |z| = z + 1 + 2i.

As per the inquiry,

We have,

    \[\left| z \right|\text{ }=\text{ }z\text{ }+\text{ }1\text{ }+\text{ }2i\]

Subbing z = x + iy, we get,

    \[\begin{array}{*{35}{l}} \Rightarrow \left| x\text{ }+\text{ }iy \right|\text{ }=\text{ }x\text{ }+\text{ }iy\text{ }+\text{ }1\text{ }+\text{ }2i \\ ~ \\ \end{array}\]

We realize that,

    \[\left| z \right|\text{ }=\text{ }\surd \left( x^2\text{ }+\text{ }y^2 \right)\]

    \[\surd \left( x^2\text{ }+\text{ }y^2 \right)\text{ }=\text{ }\left( x\text{ }+\text{ }1 \right)\text{ }+\text{ }i\left( y\text{ }+\text{ }2 \right)\]

Looking at genuine and nonexistent parts,

We get,

    \[\surd \left( x^2\text{ }+\text{ }y^2 \right)\text{ }=\text{ }\left( x\text{ }+\text{ }1 \right)\]

Also, 0 = y + 2

⇒ y = – 2

Subbing the worth of y in

    \[\surd \left( x^2\text{ }+\text{ }y^2 \right)\text{ }=\text{ }\left( x\text{ }+\text{ }1 \right),\]

We get,

    \[\Rightarrow x^2\text{ }+\text{ }\left( -\text{ }2 \right)^2\text{ }=\text{ }\left( x\text{ }+\text{ }1 \right)^2\]

    \[\Rightarrow x^2\text{ }+\text{ }4\text{ }=\text{ }x^2\text{ }+\text{ }2x\text{ }+\text{ }1\]

Consequently, x = 3/2

Consequently, z = x + iy

= 3/2 – 2i

i

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