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

Answer:

A = 2

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

Answer:

2

Explanation:

We have the expression $\frac{8-i}{3-2i}$. 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:

$\frac{8-i}{3-2i}*\frac{3+2i}{3+2i} =\frac{24-3i+16i+2}{9+4} =\frac{26+13i}{13} =2+i$

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

Hope this helps!