Answer:
−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
(
)(
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
is rewritten in the standard
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Answers & Comments
Answer:
−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