Question

Write the following expression in standard for a + bi


`(6-i)/(1+i)`
​

Answer 1

5/2-7/2 i or 2.5-3.5i

Explanation:

expand the fraction remove the parenthesis and then calculate

Answer 2

`\large\boxed{\boxed{\underline{\underline{\maltese{\pink{\pmb{\sf{\: Solution \dashrightarrow (5)/(2)-(7)/(2)i  }}}}}}}}`

Explanation:

`\frac { 6 - i } { 1 + i }\\`

Multiply the numerator & denominstor by the conjugate of 1 - i.

`(\left(6-i\right)\left(1-i\right))/(\left(1+i\right)\left(1-i\right)) \\`

We can solve the denominator by using the identity: a² - b² = (a + b) (a - b). So,

`(\left(6-i\right)\left(1-i\right))/(1^(2)-i^(2))`

We know that, i² = - 1. Then, on solving..

`(\left(6-i\right)\left(1-i\right))/(2)`

Now, simplify this expression,

`(6* 1+6\left(-i\right)-i-\left(-i^(2)\right))/(2) \\= (6* 1+6\left(-i\right)-i-\left(-\left(-1\right)\right))/(2) \\= (6-6i-i-1)/(2)`

Combine the real & imaginary parts in 6 - 6i - i - 1.

`(6-1+\left(-6-1\right)i)/(2)`

Simplify it again by adding.

`(5-7i)/(2) \\`

Now seperate the fraction by dividing 5 & -7i by 2.

`\large{\boxed{\sf(5)/(2)-(7)/(2)i }}`

__________

`\sf{Hope \: it \: helps}\\\mathfrak{Lucazz}`