Also can someone help me with equations that go like z5=32 because khan academy says the answer is 0.618+1.902i but I am so confused as to how to solve that
No, I don't believe that the Khan academy says that THE answer is 0.618+ 1.902i. Just as the quadratic equation, z
2= 4 has the
two solutions, 2 and -2, so a fifth power equation has
five roots ONE of which is (approximately) 0.618+ 1.902i. First, one obvious solution to z
5= 32 is z= 2 because 2
5= 2(2)(2)(2)(2)= 4(2)(2)(2)= 8(2)(2)= 16(2)= 32. Writing the complex number, z, in "polar form" where we write the complex number,
z=a+bi as
z=rcos(θ)+irsin(θ), or, equivalently,
z=reiθ where r and
θ are real numbers, r positive. Then
z5=r5e5iθ=32=25=25(ei0. Since r is a positive real number, r= 2.
θ=0/5=0, which gives z= 2 but since
θ is an angle, adding
2π gives the same thing. But dividing by 5 gives different values.
(0+2π)/5=(2/5)π.
cos((2/5)π)=0.3090 and
sin((2/5)π)=0.9510 so z= 2(0.3090+ 0.9510i)= 0.6180+ 1.9020i is another and that is what you are getting from Kahn academay.
A third root is given by taking
4π/5.
cos(4π/5)=−0.8090 and
sin(4π/5)=0.5878 so a third root is
z=2(−0.8090+0.5878i)=−1.6180+1.1756i.
A fourth root is given by taking
6π/5.
cos(6π/5)=−0.8090 and
sin(6π/5)=−0.5878 so a fourth root is
2(−0.08090−0.5878i)=−0.1618−1.1757i
A fifth root is given by
8π/5.
cos(8π/5)=0.3090 and
sin(8π/5)=−0.9510 so a fourth root is
2(0.3090−0.9510i)=0.6180−1.9020i.
Notice that if we tried this again, we would use
10π/5=2π which would put us back to the real root, 2 again and then the same 5 roots repeat.