Solution of equation |z|

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4-1.Complex numbers

medium

The solution of the equation $|z| - z = 1 + 2i$ is

$2 - \frac{3}{2}i$

$\frac{3}{2} + 2i$

$\frac{3}{2} - 2i$

$ - 2 + \frac{3}{2}i$

Solution

(c)$|z| – z = 1 + 2i$
Let $z = x + iy$, therefore $|x + iy| – (x + iy) = 1 + 2i$
Equating real and imaginary parts, we get
$\sqrt {{x^2} + {y^2}} – x = 1$and $y = – 2$==>$x = \frac{3}{2}$
Hence complex number $z = \frac{3}{2} – 2i$.
Trick : Since $\left| {\frac{3}{2} – 2i} \right| – \left( {\frac{3}{2} – 2i} \right)$
$ = \sqrt {\frac{9}{4} + 4} – \frac{3}{2} + 2i = \frac{5}{2} – \frac{3}{2} + 2i = 1 + 2i$

Standard 11

Mathematics

If $z_1, z_2 $ are any two complex numbers, then $|{z_1} + \sqrt {z_1^2 – z_2^2} |$ $ + |{z_1} – \sqrt {z_1^2 – z_2^2} |$ is equal to

medium

The set of all $\alpha \in R$, for which $w = \frac{{1 + \left( {1 – 8\alpha } \right)z}}{{1 – z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and ${\mathop{\rm Re}\nolimits} \,z \ne 1$, is

hard

(JEE MAIN-2018)

If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

hard

(JEE MAIN-2023)

Let $z$ be complex number satisfying $|z|^3+2 z^2+4 z-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.

Match each entry in List-$I$ to the correct entries in List-$II$.

List-$I$ List-$II$

($P$) $|z|^2$ is equal to ($1$) $12$

($Q$) $|z-\bar{z}|^2$ is equal to ($2$) $4$

($R$) $|z|^2+|z+\bar{z}|^2$ is equal to ($3$) $8$

($S$) $|z+1|^2$ is equal to ($4$) $10$

($5$) $7$

The correct option is:

normal

(IIT-2023)

Number of complex numbers $z$ such that $\left| z \right| + z – 3\bar z = 0$ is equal to