# powers of complex numbers help

#### basicange

##### New member

So I am taking a precalc course in khan academy and this lesson has to do with powers of complex numbers. I understand it until this point. Where does the 0 and negative one come from??

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

#### lookagain

##### Elite Member
View attachment 26929
So I am taking a precalc course in khan academy and this lesson has to do with powers of complex numbers. I understand it until this point. Where does the 0 and negative one come from??

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

cos(270 degrees) = 0 and sin(270 degrees) = -1

Did you learn some of that trigonometry yet?

#### HallsofIvy

##### Elite Member
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, z2= 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 z5= 32 is z= 2 because 25= 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, $$\displaystyle z= a+ bi$$ as $$\displaystyle z= r cos(\theta)+ i r sin(\theta)$$, or, equivalently, $$\displaystyle z= re^{i\theta}$$ where r and $$\displaystyle \theta$$ are real numbers, r positive. Then $$\displaystyle z^5= r^5e^{5i\theta}= 32= 2^5= 2^5(e^{i0}$$. Since r is a positive real number, r= 2. $$\displaystyle \theta= 0/5= 0$$, which gives z= 2 but since $$\displaystyle \theta$$ is an angle, adding $$\displaystyle 2\pi$$ gives the same thing. But dividing by 5 gives different values.

$$\displaystyle (0+ 2\pi)/5= (2/5)\pi$$. $$\displaystyle cos((2/5)\pi)= 0.3090$$ and $$\displaystyle sin((2/5)\pi)= 0.9510$$ 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 $$\displaystyle 4\pi/5$$. $$\displaystyle cos(4\pi/5)= -0.8090$$ and $$\displaystyle sin(4\pi/5)= 0.5878$$ so a third root is $$\displaystyle z= 2(-0.8090+ 0.5878i)= -1.6180+ 1.1756i$$.

A fourth root is given by taking $$\displaystyle 6\pi/5$$. $$\displaystyle cos(6\pi/5)= -0.8090$$ and $$\displaystyle sin(6\pi/5)= -0.5878$$ so a fourth root is $$\displaystyle 2(-0.08090- 0.5878i)= -0.1618- 1.1757i$$

A fifth root is given by $$\displaystyle 8\pi/5$$. $$\displaystyle cos(8\pi/5)= 0.3090$$ and $$\displaystyle sin(8\pi/5)= -0.9510$$ so a fourth root is $$\displaystyle 2(0.3090- 0.9510i)= 0.6180- 1.9020i$$.

Notice that if we tried this again, we would use $$\displaystyle 10\pi/5= 2\pi$$ which would put us back to the real root, 2 again and then the same 5 roots repeat.