Further Maths Complex numbers: arithmetic Practice Questions

Write down the expression

Both numbers are already in the form $a+bi$.

Group the real parts and the imaginary parts

Real parts combine with real parts, imaginary with imaginary.

State the answer in the form $a+bi$

This is the value of the expression.

Recall the definition of the complex conjugate

Only the sign of the imaginary part changes.

Identify the real and imaginary parts

Read them off from the given number.

Check that the product with the original is real

$z\overline{z}=a^{2}+b^{2}$ is always real and non-negative.

State the complex conjugate

This is the required conjugate.

Write down the quotient

The denominator must be made real before the answer can be read off.

Write down the conjugate of the denominator

The conjugate reverses the sign of the imaginary part.

Multiply numerator and denominator by that conjugate

This does not change the value of the fraction, but it will clear $i$ from the bottom.

Expand the numerator

Multiply out and keep the $i^{2}$ term for the moment.

Simplify the numerator using $i^{2}=-1$

The numerator is now a single complex number.

Expand the denominator

The cross terms in $i$ cancel because the two brackets are conjugates.

State the answer in the form $a+bi$

This is the value of the expression.

Write down the equation to be solved

There are $3$ roots in total, counted with multiplicity.

Look for a real root by the factor theorem

Since this is zero, $x-\left(1\right)$ is a factor.

Divide out the factor $x-\left(1\right)$

Algebraic division (or comparing coefficients) gives the other factor.

Identify the coefficients

Compare with the general quadratic $ax^{2}+bx+c=0$.

Calculate the discriminant

The discriminant is negative, so the roots are a complex conjugate pair.

Apply the quadratic formula

Substitute the coefficients into $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Write the square root of the negative number in terms of $i$

Use $\sqrt{-k}=i\sqrt{k}$.

Simplify the surd expression

Both parts of the numerator are divided by $2$.

Write down the two roots

The roots form a conjugate pair, as they must for real coefficients.

Check the sum of the roots

The sum of the roots of $x^{2}+bx+c=0$ is $-b$.

State the complete solution set

All $3$ roots of the equation are now known.

Write down what is given

The coefficients of $p$ are real, which is the key to the whole question.

Apply the conjugate root theorem

Non-real roots of a real polynomial always come in conjugate pairs.

Form the quadratic factor from the conjugate pair

The pair multiplies out to a quadratic with real coefficients.

Check the sum and product of that pair

These reproduce the coefficients of the quadratic factor.

Divide $p\left(x\right)$ by the quadratic factor

Algebraic division leaves the remaining factor.

Verify the factorisation by expanding

Expanding the product returns the original polynomial.

Factorise the remaining quadratic

The remaining quadratic has real roots.

Solve the remaining quadratic

These are the two real roots.

Check the root $x=-3-i$ in $p\left(x\right)$

Substituting the root into the polynomial gives zero.

Check the root $x=-1$ in $p\left(x\right)$

Substituting the root into the polynomial gives zero.

Check the root $x=2$ in $p\left(x\right)$

Substituting the root into the polynomial gives zero.

Recall the defining property of $i$

Every simplification of a complex product rests on this single fact.

Recall the definition of a complex number

$a$ is the real part and $b$ is the imaginary part of $z$.

Recall the complex conjugate

The conjugate is obtained by reversing the sign of the imaginary part.

Recall that a number times its conjugate is real

This is why multiplying by the conjugate clears $i$ from a denominator.

State the remaining roots

These are the other $3$ roots of the equation.