info@clay6.com
+91-9566306857
(9am to 6pm)
logo

Ask Questions, Get Answers

X
Want help in doing your homework? We will solve it for you. Click to know more.
 

If $z$ is a complex number having least absolute value and $|z-2+2i|=1$ then $z=$

$\begin{array}{1 1}(A)\;(2-\large\frac{1}{\sqrt 2})\normalsize (1-i)\\(B)\;(2-\large\frac{1}{\sqrt 2})\normalsize (1+i)\\(C)\;(2+\large\frac{1}{\sqrt 2})\normalsize (1-i)\\(D)\;(2+\large\frac{1}{\sqrt 2})\normalsize (1+i)\end{array} $

1 Answer

Need homework help? Click here.
$|z-2+2i|=1$ represents a circle with centre $(2,-2)$ and radius 1 in arg and plane
Direction of $z=\large\frac{1-i}{\sqrt 2}$
Magnitude of $z=\sqrt{(2)^2+(-2)^2}-1$
$\Rightarrow 2\sqrt 2-1$
$z=(2\sqrt 2-1)(\large\frac{1-i}{\sqrt 2})$$=(2-\large\frac{1}{\sqrt 2})$$(1-i)$
Hence (A) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

Related questions

...