- Home
- Standard 11
- Mathematics
4-1.Complex numbers
medium
The solution of the equation $|z| - z = 1 + 2i$ is
A
$2 - \frac{3}{2}i$
B
$\frac{3}{2} + 2i$
C
$\frac{3}{2} - 2i$
D
$ - 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
