A Level Math Preparation Tip #3: Complex numbers

Complex numbers usually come as 2 questions, 1 on the loci and 1 on algebraic properties involving Cartesian, Polar or Exponential form.

One common type of question is on the properties of argument and modulus when we multiply, divide or raise a complex number to a certain power. Recall:

\displaystyle \arg ({{z}_{1}}{{z}_{2}})=\arg ({{z}_{1}})+\arg ({{z}_{2}})

\displaystyle \arg \left( {\frac{{{{z}_{1}}}}{{{{z}_{2}}}}} \right)=\arg ({{z}_{1}})-\arg ({{z}_{2}})

\displaystyle \arg ({{z}_{1}}^{n})=n\arg ({{z}_{1}})

\displaystyle \left| {{{z}_{1}}{{z}_{2}}} \right|=\left| {{{z}_{1}}} \right|\left| {{{z}_{2}}} \right|

\displaystyle \frac{{\left| {{{z}_{1}}} \right|}}{{\left| {{{z}_{2}}} \right|}}=\frac{{\left| {{{z}_{1}}} \right|}}{{\left| {{{z}_{2}}} \right|}}

\displaystyle \left| {{{z}_{1}}^{n}} \right|={{\left| {{{z}_{1}}} \right|}^{n}}

C1

Next is the finding roots of a polynomial with unknown coefficients.

C2

One common “not smart” approach student is to write “since -2+i is a root, the conjugate -2-i is also a root”. Followed by factorising the polynomial into linear factors and compare coefficient to find “a” and “b” and then the roots.

It is easier to follow the instructions given the question. A root means when you sub z = -2+i into the polynomial you will get 0. Then compare coefficient (real part = 0, imaginary part = 0) to get “a” and “b”. This method is faster as once I know the values of “a” and “b”, I can use GC “polynomial root finder” to search for the 3 remaining roots (usually it is quite nice).

Also if the question did not mention “a” and “b” are real, it is not right to say -2-i is a root.

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