Question

8−i / 3−2i

If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a? (Note: i=√−1)

Answer 1

A = 2

Step-by-step explanation:

To rewrite

8−i / 3−2i

in the standard form a+bi, you need to multiply the numerator and denominator of

8−i / 3−2i

by the conjugate, 3+2i. This equals

( 8−i 3−2i )( 3+2i 3+2i )= 24+16i−3+(−i)(2i) (32)−(2i)2

Since i2=−1, this last fraction can be reduced simplified to

24+16i−3i+2 9−(−4) = 26+13i 13

which simplifies further to 2+i. Therefore, when

8−i / 3−2i

is rewritten in the standard form a + bi, the value of a is 2.

Answer 2

2

Step-by-step explanation:

We have the expression
`(8-i)/(3-2i)`. In order to simplify this, we need to multiply the bottom and top by the denominator's conjugate. The conjugate of a complex number a + bi is just a - bi. Here, the conjugate of 3 - 2i is 3 + 2i. So:

`(8-i)/(3-2i)*(3+2i)/(3+2i) =(24-3i+16i+2)/(9+4) =(26+13i)/(13) =2+i`

a is the real part of a complex number, so here, a = 2.

Hope this helps!