Question

If z =(1-i√3)/2(1-i) then |z| =

Answer 1

Answer:

1/✓2

Step-by-step explanation:

✓4/2✓2

2/2✓2

1/✓2

Answer 2

Given :

$\purple{\bigstar}\;\sf{z=\dfrac{1-i\sqrt{3}}{2(1-i)}}$

To find :

$\purple{\bigstar}\;\sf{|\;z\;|=?}$

Solution :

Converting z into the standard form of a Complex number

$\implies\sf{z=\dfrac{1-i\sqrt{3}}{2(1-i)}}$

$\implies\sf{z=\dfrac{1-i\sqrt{3}}{2-2i}}$

$\implies\sf{z=\dfrac{1-i\sqrt{3}}{2-2i}\times\dfrac{2+2i}{2+2i}}$

$\implies\sf{z=\dfrac{2(1)-2\sqrt{3}\;i+2i(1)-2\sqrt{3}\;i^2}{2^2-2^2\;i^2}}$

$\implies\sf{z=\dfrac{2-2\sqrt{3}\;i+2i-2\sqrt{3}\;i^2}{4-4\;i^2}}$

putting i² = - 1

$\implies\sf{z=\dfrac{2-2\sqrt{3}\;i+2i-2\sqrt{3}\;(-1)}{4-4\;(-1)}}$

$\implies\sf{z=\dfrac{2-2\sqrt{3}\;i+2i+2\sqrt{3}}{4+4}}$

$\implies\sf{z=\dfrac{(2+2\sqrt{3})+i(2-2\sqrt{3})}{8}}$

$\implies\sf{z=\dfrac{(2+2\sqrt{3})}{8}+\dfrac{(2-2\sqrt{3})}{8}\;i}$

From here,

$\bf{real\:part\:of\:z= Re(z) =\dfrac{2+2\sqrt{3}}{8}}$ and

$\bf{imaginary\:part\:of\:z=Im(z)=\dfrac{(2-2\sqrt{3})}{8}}$

Now,

Calculating | z |

$\implies\sf{|\;z\;|=\sqrt{[Re\;(z)]^2 + [Im\;(z)]^2\; }}$

$\implies\sf{|\;z\;|=\sqrt{\bigg(\dfrac{2+2\sqrt{3}}{8}\bigg)^2 +\bigg(\dfrac{2-2\sqrt{3}}{8}\bigg)^2}}$

$\implies\sf{|\;z\;|=\sqrt{\dfrac{4+12+8\sqrt{3}}{64} +\dfrac{4+12-8\sqrt{3}}{64}}}$

$\implies\sf{|\;z\;|=\sqrt{\dfrac{16+8\sqrt{3}+16-8\sqrt{3}}{64}}}$

$\implies\sf{|\;z\;|=\sqrt{\dfrac{32}{64}}}$

$\implies\sf{|\;z\;|=\sqrt{\dfrac{1}{2}}}$

$\implies\underbrace{\boxed{\boxed{\sf{|\;z\;|=\dfrac{1}{\sqrt{2}}}}}}$

Answer.