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Complex Numbers

Complex Numbers

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De Moivre's Theorem

Now that we understand the rules of how to work with complex numbers and can represent them in many different forms, we can put everything we've learned together into theorems that give us shortcuts when working with complex numbers.


These theorems follow directly from what we already know; you might already have an implicit understanding of what they say. Try to make them explicit for yourself before reading their formal statements:

Discussion

Let z=reiθ.

What are

  • z2?

  • z3?

  • z5?

Can you come up with a general rule that gives the value of zn for any exponent n?


Convert your answers into polar form, z=rcis(θ). Can you come up with a rule that gives the value of zn in this form?

When we multiply complex numbers, we multiply their moduli and add their arguments. In exponential form this is just the law of exponents, (ea)n=ean. Thus

z2z3z5​=(reiθ)2=r2ei(θ+θ)=r2ei(2θ)=(reiθ)3=r3ei(θ+θ+θ)=r3ei(3θ)=(reiθ)5=r5ei(5θ)​

More generally, multiplying z by itself n times gives

zn=(reiθ)n=rnei(nθ)

To convert back into the “cis” or polar‐angle form, we use

eiθ=cosθ+isinθ

so

zn=rn(cos(nθ)+isin(nθ))=rncis(nθ)

In words: each time you multiply by z you scale the length by r and rotate by θ; doing it n times scales by rn and rotates by nθ.

This is De Moivre's Theorem about powers of complex numbers.

Powers of complex numbers

Now that we know how to represent complex numbers in the form reiθ, we can understand De Moivre's Theorem, which gives us a shortcut to finding powers of complex numbers:

z=reiθ⇒zn=(reiθ)n=rneinθ


And since reiθ=rcisθ:

[r(cosθ+isinθ)]n=(rcisθ)n=rneinθ=rncis(nθ)📖

Example

Find (1−i)6.


First express 1−i in polar form:

1−i=√2⋅cis(−4π​)

So

(1−i)6  ​=(√2)6⋅cis(−46π​) =8cis(−23π​)=8i​

Since taking the nth root of any number is the same thing as raising it to the power of 1/n, we can also express De Moivre's Theorem in terms of roots.

Roots of complex numbers

De Moivre's Theorem can also be used to find the nth roots of complex numbers:

n√reiθ=(reiθ)1/n=n√r⋅eiθ/n🚫

or equivalently

n√rcisθ=n√rcis(nθ​)🚫


However, since cisθ=cis(θ+2kπ) for any k∈Z, then we actually have

n√rcisθ=n√rcis(nθ+2kπ​),k=0,1…n−1🚫

Note that k stops at n−1 since when k=n we have

cis(nθ+2nπ​)=cis(θ+2π)=cisθ

Example

Find all possible roots z∈C of the equation z4+4=0.


Rearranging we have

z4=−4=4cis(π)

So

z    ​=4√4cis(4π+2kπ​),k=0,1,2,3 =√2⋅cis({4π​,43π​,45π​ or 47π​}) =√2⋅√21​⋅{1+i,−1+i,−1−i,1−i}​


So z=1+i,−1+i,−1−i or 1−i eg z=±1±i.

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