Question

Q. If z = ((1 + i)(sqrt(3) - i))/(sqrt(3) + i) modulus of z is then the​

Answer 1

Answer:

Step-by-step explanation:

We have,

$z=\dfrac{\left(1+i\right)\left(\sqrt{3}-i\right)}{\sqrt{3}+i}$

Multiplying the denominator with its conjugate, also with the numerator,

$\implies\,z=\dfrac{\left(1+i\right)\left(\sqrt{3}-i\right)\left(\sqrt{3}-i\right)}{\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)}$

$\implies\,z=\dfrac{\left(1+i\right)\left(\sqrt{3}-i\right)^{2}}{\left(\sqrt{3}\right)^{2}+\left(1\right)^{2}}$

$\implies\,z=\dfrac{\left(1+i\right)\left\{\left(\sqrt{3}\right)^{2}-2\cdot\sqrt{3}\cdot\,i+\left(i\right)^{2}\right\}}{3+1}$

$\implies\,z=\dfrac{\left(1+i\right)\left(3-2\sqrt{3}\,i-1\right)}{4}$

$\implies\,z=\dfrac{\left(1+i\right)\left(2-2\sqrt{3}\,i\right)}{4}$

$\implies\,z=\dfrac{2\left(1+i\right)\left(1-\sqrt{3}\,i\right)}{4}$

$\implies\,z=\dfrac{\left(1+i\right)\left(1-\sqrt{3}\,i\right)}{2}$

$\implies\,z=\dfrac{1-\sqrt{3}\,i+i-\sqrt{3}\,{i}^{2}}{2}$

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

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

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

$\implies\,z=\dfrac{4+\left(1-\sqrt{3}\right)i}{2}$

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

Now,

$\implies\,|z|=\left|2+\dfrac{1-\sqrt{3}}{2}\,i\right|$

$\implies\,|z|=\left(2\right)^{2}+\left(\dfrac{1-\sqrt{3}}{2}\right)^{2}$

$\implies\,|z|=4+\dfrac{\left(1\right)^{2}+\left(\sqrt{3}\right)^{2}-2\cdot\sqrt{3}\cdot1}{4}$

$\implies\,|z|=4+\dfrac{1+3-2\sqrt{3}}{4}$

$\implies\,|z|=4+\dfrac{4-2\sqrt{3}}{4}$

$\implies\,|z|=\dfrac{16+4-2\sqrt{3}}{4}$

$\implies\,|z|=\dfrac{20-2\sqrt{3}}{4}$

$\implies\,|z|=\dfrac{10-\sqrt{3}}{2}$