Question

z^2=i. Solve in a+bi form. I have been teaching myself the complex numbers material and I am not sure how to do this.

Answer 1

there are two solutions to this. draw the unit circle in the complex plane. i sits right up top at an angle of
\[\frac{\pi}{2}\]
half that angle is
\[\frac{\pi}{4}\] so find the coordinates on the unit circle with angle of
\[\frac{\pi}{4}\] and that is your first answer

Answer 2

second answer is directly across from the first answer on the unit circle

Answer 3

at an angle of
\[\frac{\pi}{4}\] you get
\[\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\] and that is one solution

Answer 4

Hmm confusing. Is this using eulers formula?

Answer 5

i guess

Answer 6

the idea is this. when you square a complex number you square the absolute value and double the angle.

Answer 7

so if you want the square root, you work backwards. now the absolute value of i is just 1, therefore you are working on the unit circle. so you can pretty much visualize the answer. half way between 1 and i on the unit circle is the point i drew

Answer 8

Oh, okay. I was looking at notes and I didn't realize those brackets were still absolute values. Abs(z)=r=1

Answer 9

right. and there is another one as well, the one right across the circle. since you are taking square roots there will be two answers.

Answer 10

hm r will remain 1 but how did you get to a+bi

Answer 11

i suppose you could say
\[i=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\] and then divide the angle by 2 to get your answer

Answer 12

you have to know the point on the unit circle whose angle is
\[\frac{\pi}{4}\]

Answer 13

yeah got that

Answer 14

there is no way to do this without the trig i don't think. and the other one is right across at
\[-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]

Answer 15

hah im sorry but still lost. I know that i=Cos(pi/2)+isin(pi/2)

Answer 16

ok if you know that you are almost done. you want the two square roots, so divide the angle by 2

Answer 17

right, but why do you just divide by two. Why wouldn't it be the sqrt(Pi/2)

Answer 18

you get
\[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\]

Answer 19

oh because when you multiply two complex numbers together you
1) multiply the modulus (absolute value)
2) ADD the angles

Answer 20

and therefore if you want to raise a complex number to a power you
1) raise the absolute value to the power and
2) MULTIPLY the angles (not raise the angles to a power)

Answer 21

and finally if you want to find the roots of a complex number you
1) take the root of the absolute value and
2) DIVIDE the angle

Answer 22

gonna write that down! Well I think I almost got it. I am gonna think it over more. I REALLY appreciate your time and help.

Answer 23

btw you can check by computing that
\[(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^2=i\]

Answer 24

might be worth the arithmetic to see how it works out.

Answer 25

yw

Answer 26

mathematica just spit out a simplified thing. So you mean do it by hand?

Answer 27

yes i mean write
\[(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\]

Answer 28

multiply out and see what you get

Answer 29

okay I will do that but couldn't you follow your rule above and do Abs(z1)(Abs(z2)=r*r and do Pi/4+Pi/4?

Answer 30

well sure but that is how we got it to begin with.

Answer 31

just verifies that you divided by two and took the square root of 1

Answer 32

This stuff is crazy.

Answer 33

I have to tackle ln(3+4i) next if you aren't sick of this yet.

Answer 34

we wrote
\[i=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\] and took half the angle. so it stands to reason if you double the angle we took half of you will get back to what you started with. i meant multiply out in standard
\[a+bi\] form to see that it actually works.

here is a nice picture from wolfram solving
\[z^6=1\] you can see that the solutions are evenly spaced about the unit circle
http://www.wolframalpha.com/input/?i=z^6%3D1

Answer 35

So the power determines the number of solutions?

Answer 36

yes. fundamental theorem of algebra says a poly of degree n has n solution. if you want to solve
\[z^n=1\] there will be n distinct solution, evenly spaced about the unit circle

Answer 37

as for Log, it is
\[\log(z) = \ln(|z|)+i \theta\]

Answer 38

\[|3+4i|=5\] by pythagoras, the angle i don't know. so i guess you could write
\[\log(3+4i)=\ln(5)+i\tan^{-1}(\frac{4}{3})\] or you could actually compute the angle with a calculator