Question

Write the following expression in standard for a + bi

$\frac{6-i}{1+i}$
​

Answer 1

Answer:

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

Step-by-step explanation:

expand the fraction remove the parenthesis and then calculate

Answer 2

Answer:

$\large\boxed{\boxed{\underline{\underline{\maltese{\pink{\pmb{\sf{\: Solution \dashrightarrow \frac{5}{2}-\frac{7}{2}i  }}}}}}}}$

Step-by-step explanation:

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

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

$\frac{\left(6-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$

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

$\frac{\left(6-i\right)\left(1-i\right)}{1^{2}-i^{2}}$

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

$\frac{\left(6-i\right)\left(1-i\right)}{2}$

Now, simplify this expression,

$\frac{6\times 1+6\left(-i\right)-i-\left(-i^{2}\right)}{2} \\= \frac{6\times 1+6\left(-i\right)-i-\left(-\left(-1\right)\right)}{2} \\= \frac{6-6i-i-1}{2}$

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

$\frac{6-1+\left(-6-1\right)i}{2}$

Simplify it again by adding.

$\frac{5-7i}{2}$

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

$\large{\boxed{\sf\frac{5}{2}-\frac{7}{2}i }}$

__________

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