Using De Moivre's Theorem to solve equations HELP

[itsangelan]

Hi guys, im currently having difficulty just understanding this chapter so can someone please help me with the following questions. Thank you

For each of the following, solve the equation over C and show the solutions on an Argand diagram.
a) z^2 + 1 = 0
b) z^3 = 27i
c) z^2 = 1 + root 3i

[de]

Hi guys, im currently having difficulty just understanding this chapter so can someone please help me with the following questions. Thank you

For each of the following, solve the equation over C and show the solutions on an Argand diagram.
a) z^2 + 1 = 0
b) z^3 = 27i
c) z^2 = 1 + root 3i

So the trick with these is to express z as . Then you get in a)
cis always has magnitude one. So r=1 here.
so and here you use demoivre so So for integer n and which gives you points on an argand diagram at i and -i.
With b you'll need r to be 3 of course c) is a little trickier but see what you can do (thinking about what the RHS might look like in cis format)

[zhen]

Hi guys, im currently having difficulty just understanding this chapter so can someone please help me with the following questions. Thank you

For each of the following, solve the equation over C and show the solutions on an Argand diagram.
a) z^2 + 1 = 0
b) z^3 = 27i
c) z^2 = 1 + root 3i

a) z^2=-1
On the argand diagram, it helps to think of the real axis as the x-axis and the imaginary axis as the y-axis. So, -1 is π or 180° around.
z^2=cis(π+2πk), where k is an integer (this is because every 2π, you go around the circle back to the original place.
z=cis(π/2+πk)
Let k=0
z=cis(π/2)
Let k=-1
z=cis(-π/2)

b) z^3=27i
z^3=27cis(π/2+2πk), where k is an integer
z=3cis(π/6+2πk/3)
Note: When you let k=a number, we choose numbers to find all the possible answers, but it doesn't matter what number we choose, which is why we choose easy numbers to sub in like 0 and 1. For this since it's z^3, there are 3 different solutions so sub in 3 different numbers into k.

c) z^2=1 + root 3 i
To find the polar form of this we need to find the modulus and argument, where rcis(x), where r is the modulus and x is the argument.

To find the modulus, of z^2=a+bi you do the root of a^2+b^2, so in this question the modulus is 2.

To find the argument you do argtan or inverse tan the b/a, when z^2=a+bi. tan^-1(root 3/1) =π/3
Note: If it was z=1-i, the argument would be -π/4, as this is in the fourth quadrant because of the negative in front of i. If it was z=-1+i, the argument is 3π/4, as it is in the second quadrant.

So it's z^2=2cis(π/3+2πk), where k is an integer

To finish this just sub in values for k

I haven't turned it into the rectangular form, but if you want to cis(x)=cos(x)+isin(x)

My answers might be wrong, as I might have made a mistake, so someone should double check, but the process should be ok.

[itsangelan]

Hi zhen, Im a little slow so please bear with me. For a, where did you get the equation cis(pi + 2kpi) from?

[itsangelan]

Oh nevermind i got it, thankyou bothn