Complex Numbers | IB Math AAHL | Perplex

Complex Numbers

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Watch comprehensive video reviews for Complex Numbers, designed for final exam preparation. Each video includes integrated problems you can solve alongside detailed solutions.

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Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

Not your average video:

Interactive Problems: Solve problems alongside the video with step-by-step guidance and detailed solutions.

Exam Preparation: Complete unit reviews designed for final exam preparation with all key concepts covered systematically.

Expert Teaching: High-quality instruction from Perplex co-founder James Mullen with clear explanations, worked examples, and exam tips.

Cartesian formPolar formDe Moivre

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Roots of complex numbers

AHL AA 1.14

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.

Roots of complex numbers

AHL AA 1.14

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.