Question

Express the complex number z=8÷1+i√3 in a+ib form

Answer 1

Solution

z = 2 - 2√3

Given

$\sf \: z = \dfrac{8}{1 + \sqrt{3} \iota }$

To express z in a + in form

Now,

Multiplying and Dividing by the conjugate of the denominator

$\longrightarrow \: \sf \: z = \dfrac{8}{1 + \sqrt{3} \iota } \times \dfrac{1 - \sqrt{3} \iota}{1 - \sqrt{3} \iota } \\ \\ \longrightarrow \: \sf \: z = \dfrac{8(1 - \sqrt{3} \iota) }{ {1}^{2} - ( \sqrt{3} \iota) {}^{2} } \\ \\ \longrightarrow \ \sf \: z = \dfrac{8 - 8 \sqrt{3} \iota}{1 - 3( - 1)} \\ \\ \longrightarrow \: \sf \: z = \dfrac{8 - 8 \sqrt{3} \iota}{4} \\ \\ \longrightarrow \: \sf \: z = 2 - 2 \sqrt{3} \iota$

On comparing with z = a + ib,

a = 2 and b = -2√3

Note

• I is referred to as Iota or Imaginary Number

• For instance,we are asked to find the roots of √4 we would say ±2 but if it is √-4. Here,we assume √- 1 as I and rewrite the equation as √4i whose roots are ± 2i

• I² = I × I = (√-1)(√-1) = (√-1)² = - 1

Answer 2

$\huge{\underline{\underline{\green{\mathfrak{AnSwEr :}}}}}$

We are given that,

$\large{\boxed{\sf{z \: = \: \dfrac{8}{1 \: + \: \sqrt{3} \iota}}}} \\ \\ {\small{\underline{\gray{\sf{\dag \: \: \: \: \: \: Rationalize \: the \: Denominator \: \: \: \: \: \: \dag}}}}} \\ \\ \implies{\sf{z \: = \: \dfrac{8}{1 \: + \: \sqrt{3} \iota} \: \times \: \dfrac{1 \: - \: \sqrt{3} \iota}{1 \: - \: \sqrt{3} \iota}}} \\ \\ \implies {\sf{z \: = \: \dfrac{8(1 \: - \: \sqrt{3} \iota)}{1^2 - (\sqrt{3} \iota)^2}}} \\ \\ \implies {\sf{z \: = \: \dfrac{8 - 8\sqrt{3} \iota}{1 - 3( - 1)}}} \\ \\ \sf{\implies z \: = \: \dfrac{8 - 8 \sqrt{3} \iota}{1 + 3}} \\ \\ \sf{\implies z \: = \: \dfrac{8 - 8\sqrt{3} \iota}{4}} \\ \\ \sf{\implies z \: = \: \dfrac{\cancel{4}(2 - 2\sqrt{3} \iota)}{\cancel{4}}} \\ \\ \sf{\implies z \: = \: 2 - 2\sqrt{3} \iota}$

So,

• Real Part (a) = 2
• Imaginary part (ib) = $\sf{-2\sqrt{3} \iota}$