Question

express the complex number z=8/1+i root3 in a+ib form
$8 \div 1 + i \sqrt{3 \: in \: a + ib \: form}$
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Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{z=\dfrac{8}{1+i\sqrt3}}$

$\underline{\textbf{To find:}}$

$\mathsf{a+i\,b\;form\;of\;z=\dfrac{8}{1+i\sqrt3}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{z=\dfrac{8}{1+i\sqrt3}}$

$\textsf{Multiply both numerator and denominator by}$

$\mathsf{conjugate\;of\;1+i\sqrt3}$

$\mathsf{z=\dfrac{8}{1+i\sqrt3}{\times}\dfrac{1-i\sqrt3}{1-i\sqrt3}}$

$\mathsf{z=\dfrac{8(1-i\sqrt3)}{(1+i\sqrt3)(1-i\sqrt3)}}$

$\textsf{Using the identity,}\;\;\boxed{\bf\,(a+b)(a-b)=a^2-b^2}$

$\mathsf{z=\dfrac{8(1-i\sqrt3)}{1^2-i^2(\sqrt3)^2}}$

$\mathsf{z=\dfrac{8(1-i\sqrt3)}{1+3}}$

$\mathsf{z=\dfrac{8(1-i\sqrt3)}{4}}$

$\mathsf{z=2(1-i\sqrt3)}$

$\boxed{\bf\,z=2-i\,2\sqrt3}$

$\textsf{which is of the form a+ib}$

Answer 2

Answer:

The a+ib form of  $z=\frac{8}{1+i\sqrt{3} }$  is  $z=2-i2\sqrt{3}$

Step-by-step explanation:

Rationalize:

• Given complex number is

        $z=\frac{8}{1+i\sqrt{3} }$

• Rationalize the above number

          Multiply and divide the number with ${1-i\sqrt{3} }$

                  $z=\frac{8}{1+i\sqrt{3} }(\frac{1-i\sqrt{3} }{1-i\sqrt{3} })$

                     $=\frac{8(1-i\sqrt{3}) }{1+3}$

                     $=2(1-i\sqrt{3})$

                  $z=2-i2\sqrt{3}$

        Hence, a+ib form $z=\frac{8}{1+i\sqrt{3} }$  is  $z=2-i2\sqrt{3}$  

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