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
a would be [(-3/sqrt{2})+i*(3/sqrt{2})\]
b would be -1+i*sqrt(3)
seems unlikely
first for a and second for b
sorry b willbe -1 -i*3/sqrt(2)
\[3(\cos(27°) + i\sin(27°))^5\] raise \(3^5\) to get \(343\) then multiply the angle by \(5\) to get \[27\times 5=135\] so it is going to be \[343\left(\cos(135)+i\sin(135)\right)\]
then evaluate the trig functions \[\cos(135)=-\frac{\sqrt2}{2}, \sin(135)=\frac{\sqrt2}{2}\] and plug them in
3 is out of شقؤ
3 is out of arc so i still 3
\[343\left(\cos(135)+i\sin(135)\right)\] \[=343\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)\]
So what Satellite said what the correct answer for a? Sorry I am a little bit confused
sqrt(2)/2 =1/sqrt(2)
the power is on the arc only but 3 is free so 3 still 3
i really doubt it
a will be [(-3/sqrt{2})+i*(3/sqrt{2})\]
b will be -1 -i*3/sqrt(2)
i think it is the right answer
I will try it :)
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