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OpenStudy (anonymous):
Please help! Perform the indicated operation. Leave all answers in the form a + b i.
(1 - i ) (2 + 3 i )
a. -1 - 5 i
b. -1 + i
c. 5 + i
d. 5 - 5 i
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hartnn (hartnn):
do u know this:
(a+b)(c+d)=ac+ad+bc+bd
?
OpenStudy (anonymous):
I'm familiar. Hmmm, so that's what I use?
hartnn (hartnn):
and also the fact that i*i = -1
just try it....
mathslover (mathslover):
\[\large{(1-i)(2+3i)=1(2+3i)-i(2+3i)}\]
\[\large{(1-i)(2+3i)=1(2)+1(3i)-i(2)-i(3i)}\]
this is the use of taht @IloveCharlie
mathslover (mathslover):
*that
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OpenStudy (anonymous):
Hmm... :/
OpenStudy (anonymous):
d.
mathslover (mathslover):
what is i(3i) ?
mathslover (mathslover):
i * 3i = 3 i ^2 = 3(-1) = -3
2+3i-2i+3
5+i
OpenStudy (anonymous):
i(3i)=-3 @mathslover
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OpenStudy (anonymous):
9+2i?
OpenStudy (anonymous):
:/
hartnn (hartnn):
@mathslover already gave u answer,see the comments....
OpenStudy (anonymous):
Yes but I can't match it up :/
hartnn (hartnn):
he gave u answer as 5+i,which is option c
here are all steps together:
(1−i)(2+3i)
=1(2)+1(3i)−i(2)−i(3i)
=2+3i-2i+3
=5+i
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OpenStudy (anonymous):
Oh I understood I just must have missed it. Thank you for explaining the steps BOTH of you I appreciate it
mathslover (mathslover):
Oh sorry yes :
i *3 i =3 i^2 = -3
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