Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

8. Write each expression in the standard form for a complex number, a + bi. a. [3(cos(27°)) + isin(27°)]5 b. [2(cos(40°)) + isin(40°)]6

OpenStudy (anonymous):

a would be [(-3/sqrt{2})+i*(3/sqrt{2})\]

OpenStudy (anonymous):

b would be -1+i*sqrt(3)

OpenStudy (anonymous):

seems unlikely

OpenStudy (anonymous):

first for a and second for b

OpenStudy (anonymous):

sorry b willbe -1 -i*3/sqrt(2)

OpenStudy (anonymous):

\[3(\cos(27°) + i\sin(27°))^5\] raise \(3^5\) to get \(343\) then multiply the angle by \(5\) to get \[27\times 5=135\] so it is going to be \[343\left(\cos(135)+i\sin(135)\right)\]

OpenStudy (anonymous):

then evaluate the trig functions \[\cos(135)=-\frac{\sqrt2}{2}, \sin(135)=\frac{\sqrt2}{2}\] and plug them in

OpenStudy (anonymous):

3 is out of شقؤ

OpenStudy (anonymous):

3 is out of arc so i still 3

OpenStudy (anonymous):

\[343\left(\cos(135)+i\sin(135)\right)\] \[=343\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)\]

OpenStudy (anonymous):

So what Satellite said what the correct answer for a? Sorry I am a little bit confused

OpenStudy (anonymous):

sqrt(2)/2 =1/sqrt(2)

OpenStudy (anonymous):

the power is on the arc only but 3 is free so 3 still 3

OpenStudy (anonymous):

i really doubt it

OpenStudy (anonymous):

a will be [(-3/sqrt{2})+i*(3/sqrt{2})\]

OpenStudy (anonymous):

b will be -1 -i*3/sqrt(2)

OpenStudy (anonymous):

i think it is the right answer

OpenStudy (anonymous):

I will try it :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!