Question:
If $z$ satisfies the equation $|z|-z=1+2 i$, then $z$ is equal to
If $z$ satisfies the equation $|z|-z=1+2 i$, then $z$ is equal to
Updated On: Apr 15, 2024
- $\frac{3}{2} + 2i$
- $\frac{3}{2} - 2i$
- $ 2 - \frac{3}{2} i $
- $ 2 + \frac{3}{2} i $
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The Correct Option is B
Solution and Explanation
We have, $|z|-z=1+2 i$
If $z=x +i y$, then this equation reduces to
$|x +i y|-(x +i y)=1+2 i$
$\Rightarrow \left(\sqrt{x^{2}+y^{2}}-x\right)+(-i y)=1+2 i$
On comparing real and imaginary parts of both sides of this equation, we get $\sqrt{x^{2}+y^{2}}-x=1$
$\Rightarrow \sqrt{x^{2}+y^{2}}=1+x$
$\Rightarrow x^{2}+y^{2}=(1+x)^{2}$
$\Rightarrow x^{2}+y^{2}=1+x^{2}+2 x$
$\Rightarrow y^{2}=1+2 x$ ...(i)
and $-y=2$
$\Rightarrow y=-2$
Putting this value in E (i), we get
$(-2)^{2}=1+2 x$
$\Rightarrow 2 x=3$
$\Rightarrow x =\frac{3}{2}$
Hence, $z =x +i y$
$=\frac{3}{2}-2 i $
If $z=x +i y$, then this equation reduces to
$|x +i y|-(x +i y)=1+2 i$
$\Rightarrow \left(\sqrt{x^{2}+y^{2}}-x\right)+(-i y)=1+2 i$
On comparing real and imaginary parts of both sides of this equation, we get $\sqrt{x^{2}+y^{2}}-x=1$
$\Rightarrow \sqrt{x^{2}+y^{2}}=1+x$
$\Rightarrow x^{2}+y^{2}=(1+x)^{2}$
$\Rightarrow x^{2}+y^{2}=1+x^{2}+2 x$
$\Rightarrow y^{2}=1+2 x$ ...(i)
and $-y=2$
$\Rightarrow y=-2$
Putting this value in E (i), we get
$(-2)^{2}=1+2 x$
$\Rightarrow 2 x=3$
$\Rightarrow x =\frac{3}{2}$
Hence, $z =x +i y$
$=\frac{3}{2}-2 i $
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.