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Complex number in standard form. How do I do this? 5/(1+i)
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is a+bi standard form? if so you might want to run a conjugate
Yes
then a conjugate it is: recall that (a+b)(a-b)=a^2-b^2
ok now what do i do?
essentially you mulitply by 1, but in the form of the conjugate ....
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\[\frac{k}{a+b}*1\] \[\frac{k}{a+b}*\frac{a-b}{a-b}\] \[\frac{ka-kb}{a^2-b^2}\] \[\frac{ka}{a^2-b^2}-\frac{kb}{a^2-b^2}\]
What is k?
It's just a general variable. Here's what you'll be doing. Take your denominator and flip the sign between the 1 and the i. This is your conjugate, in this case it's (1-i). Now multiply the numerator and denominator by (1+i)
(5-5i)/2
Right?
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Right, exactly, so thats your answer: 2.5 - 2.5i
Thank you so much!
No problem, glad I could help
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