Operations with Complex Numbers (DP IB Applications & Interpretation (AI): HL): Revision Note

Complex addition, subtraction & multiplication

How do I add and subtract complex numbers?

• To add or subtract complex numbers, add or subtract their real and imaginary parts separately

  • e.g.

    • $\left(3+2\mathrm{i}\right)+\left(-4+\mathrm{i}\right)=-1+3\mathrm{i}$

    • $\left(3+2\mathrm{i}\right)-\left(-4+\mathrm{i}\right)=7+\mathrm{i}$

How do I multiply complex numbers?

• To multiply $z_1=a+b\mathrm{i}$ by $z_2=c+d\mathrm{i}$ you need to expand brackets

  • $z_1{z}_2=\left(a+b\mathrm{i}\right)\left(c+d\mathrm{i}\right)$

    • just remember that ${\mathrm{i}}^2=-1$

• e.g. $\left(2+3\mathrm{i}\right)\left(4+5\mathrm{i}\right)=8+10\mathrm{i}+12\mathrm{i}+15{\mathrm{i}}^2$

  • ${\mathrm{i}}^2=-1$ gives $8+10\mathrm{i}+12\mathrm{i}-15$

  • which simplifies to $-7+22\mathrm{i}$

How do I find powers of i?

• Because ${\mathrm{i}}^2=-1$, higher powers of $\mathrm{i}$ can be found as follows:

  • ${\mathrm{i}}^3={\mathrm{i}}^2\times \mathrm{i}=\mathrm{ }-\mathrm{i}$

  • ${\mathrm{i}}^4={\left({\mathrm{i}}^2\right)}^2={\left(-1\right)}^2=1$

  • ${\mathrm{i}}^5={\left({\mathrm{i}}^2\right)}^2\mathrm{ }\times \mathrm{i}=\mathrm{i}$

  • ${\mathrm{i}}^6={\left({\mathrm{i}}^2\right)}^3={\left(-1\right)}^3=\mathrm{ }-1$

• The powers of $\mathrm{i}$ form a sequence with period 4

  • $\mathrm{i},-1,-\mathrm{i},1,\mathrm{i},-1,-\mathrm{i},1,...$

• Use index laws to find much higher powers

  • ${\mathrm{i}}^{23}={\left({\mathrm{i}}^2\right)}^{11}\times \mathrm{i}={\left(-1\right)}^{11}\times \mathrm{i}=\mathrm{ }-\mathrm{i}$

  • Just remember that

    • ${\left(-1\right)}^n=1$ if $n$ is even

    • ${\left(-1\right)}^n=-1$ if $n$ is odd

Complex conjugation & division

What is a complex conjugate?

• For the complex number $z=a+b\mathrm{i}$, the complex conjugate of $z$ (written $z^{*}$) is

  • $z^{*}=a-b\mathrm{i}$

• If $z=a-b\mathrm{i}$ then $z^{*}=a+b\mathrm{i}$

• You will find that:

  • $z+z^{*}$ is always real

    • e.g. $(6+5\mathrm{i})+(6-5\mathrm{i})=6+6+5\mathrm{i}-5\mathrm{i}=12$

  • $z-z^{*}$ is always imaginary

    • e.g. $(6+5\mathrm{i})-(6-5\mathrm{i})=6-6+5\mathrm{i}-(-5\mathrm{i})=10\mathrm{i}$

  • $z{z}^{*}$ or $z^{*}z$ is always real

    • e.g. $(6+5\mathrm{i})(6-5\mathrm{i})=36+30\mathrm{i}–30\mathrm{i}-25{\mathrm{i}}^2=36–25(-1)=61$

How do I divide complex numbers?

• To divide two complex numbers, do the following:

  • STEP 1
    Write the division as a fraction

  • STEP 2
    Multiply the top and bottom of the fraction by the conjugate of the denominator

    • $\frac{a+b\mathrm{i}}{c+d\mathrm{i}}=\frac{a+b\mathrm{i}}{c+d\mathrm{i}}\times \frac{c-d\mathrm{i}}{c-d\mathrm{i}}$

    • This is similar to rationalising a denominator with surds

  • STEP 3
    Add in brackets and multiply out the top and bottom, simplifying your answer

    • The denominator will always be real

  • STEP 4
    Give your answer in the form required

    • or in Cartesian form $p+q\mathrm{i}$ if not specified