What does a negative discriminant tell you about the roots of a quadratic with real coefficients?

The quadratic has no real roots , so its graph never crosses the horizontal axis. Instead it has two complex roots, which come from taking the square root of a negative number in the quadratic formula.

True or False? The complex roots of any quadratic equation always occur as a conjugate pair.

False. This holds only when the coefficients of the quadratic are real. If a coefficient is itself complex the two roots need not be conjugates, and the quadratic formula has to be used to find them.

Why must a cubic equation with real coefficients always have at least one real root?

It has three roots in total, and non-real roots always arrive in conjugate pairs , so the number of non-real roots has to be even. An even number cannot make up three on its own, so at least one root is left over and must be real.

You are given one complex root of a cubic with real coefficients. What is the quickest route to the other two roots?

Write down the conjugate as a second root, then multiply the two factors together to get a real quadratic factor . Dividing the cubic by that quadratic, or comparing coefficients, leaves the linear factor and so the third root.

A polynomial equation has unknown real coefficients and you are given one complex root. What do you do with the root?

Substitute it in, expand, and collect the real and imaginary parts separately. Comparing parts then gives simultaneous equations in the unknown coefficients, and once those are known the remaining roots are found in the usual way.

Which complex numbers have a negative principal argument?

Those whose imaginary part is negative , which is the third and fourth quadrants of an Argand diagram. Turning clockwise from the positive real axis reaches such a point sooner than turning anticlockwise, so the principal argument is recorded as negative.

A LevelFurther MathsEdexcelCore PureFlashcardsComplex NumbersComplex Numbers & Argand Diagrams

Complex Numbers & Argand Diagrams (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontIntroduction to Complex Numbers
Define the imaginary unit $i$ used to write complex numbers.

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Cards in this collection (52)

Define the imaginary unit $i$ used to write complex numbers.
$i$ is defined so that $i^{2} = - 1$ , which makes it a square root of $- 1$ .
It lets the square root of any negative number be written as a real multiple of $i$ , since $\sqrt{- a} = \sqrt{a} \times \sqrt{- 1}$ .
For the complex number $z = 3 - 7 i$ write down its real part and its imaginary part.
$Re (z) = 3$ and $Im (z) = - 7$ .
The imaginary part is the real coefficient of $i$ , so it does not include the $i$ itself and it keeps the minus sign.
True or False?
$\sqrt{- 4} \times \sqrt{- 9} = \sqrt{36} = 6$
False.
Each root has to be written as a multiple of $i$ first, giving $\sqrt{- 4} \times \sqrt{- 9} = 2 i \times 3 i = 6 i^{2} = - 6$ .
The surd rule $\sqrt{a} \times \sqrt{b} = \sqrt{a b}$ holds only when $a$ and $b$ are not both negative.
Fill in the missing powers of $i$ :
$i^{2} = - 1, so i^{3} =______, i^{4} =______and i^{5} =______$
The completed line is:
$i^{2} = - 1, so i^{3} = - i, i^{4} = 1 and i^{5} = i$
The powers of $i$ repeat in a cycle of four, so once you reach $i^{4}$ the pattern begins again.
How do you evaluate a high power of $i$ such as $i^{23}$ quickly?
Split off the largest even power and use $i^{2} = - 1$ , which gives $i^{23} = {(i^{2})}^{11} \times i$ .
Since ${(- 1)}^{11} = - 1$ , this comes out as $i^{23} = - i$ .
When you expand $(3 + 4 i) (6 + 7 i)$ which term needs extra care?
The $i \times i$ term is the one to watch.
Expanding gives $18 + 21 i + 24 i + 28 i^{2}$ , and replacing $i^{2}$ by $- 1$ turns $28 i^{2}$ into the real number $- 28$ .
Collecting up then leaves $- 10 + 45 i$ .
Define the complex conjugate $z^{*}$ of a complex number.
For $z = a + b i$ the complex conjugate is $z^{*} = a - b i$ : the real part is unchanged and the sign of the imaginary part is reversed.
So if $z = 6 + 5 i$ then $z^{*} = 6 - 5 i$ .
Why does a single equation between two complex numbers give you two real equations?
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.
So an equation such as $a + b i = 5 - 2 i$ splits into the two real equations $a = 5$ and $b = - 2$ .
For $z = a + b i$ complete these results about $z$ and its conjugate:
$z + z^{*} =______, z - z^{*} =______and z z^{*} =______$
The completed results are:
$z + z^{*} = 2 a, z - z^{*} = 2 b i and z z^{*} = a^{2} + b^{2}$
So $z + z^{*}$ and $z z^{*}$ are always real, while $z - z^{*}$ is always imaginary.
When dividing by a complex number, what do you multiply the top and bottom of the fraction by?
Multiply both by the conjugate of the denominator.
A complex number multiplied by its conjugate is always real, so the denominator becomes a real number and what is left is a division by a real number.
Multiplying top and bottom by the same thing is multiplying by $1$ , so the value of the fraction is unchanged.
Why should the complex number $3 \sqrt{5} i$ be written as $3 i \sqrt{5}$ instead?
Written as $3 \sqrt{5} i$ the $i$ sits immediately after the surd, so it can be misread as $3 \sqrt{5 i}$ with the $i$ inside the root.
Putting the $i$ in front of the surd, or bracketing the real part as $(3 \sqrt{5}) i$ , makes clear that only $5$ is under the root.
What does a negative discriminant tell you about the roots of a quadratic with real coefficients?
The quadratic has no real roots, so its graph never crosses the horizontal axis.
Instead it has two complex roots, which come from taking the square root of a negative number in the quadratic formula.
A quadratic equation with real coefficients has $2 - 3 i$ as one of its roots. What is the equation?
The other root must be the conjugate $2 + 3 i$ , so $z - (2 - 3 i)$ and $z - (2 + 3 i)$ are both factors.
Multiplying them out gives the equation $z^{2} - 4 z + 13 = 0$ .
Complete the expansion that shows why a conjugate pair of roots gives a real quadratic factor:
$(z - (m + n i)) (z - (m - n i)) = {(______)}^{2} +______$
The completed expansion is:
$(z - (m + n i)) (z - (m - n i)) = {(z - m)}^{2} + n^{2}$
Regrouping the factors as $((z - m) - n i) ((z - m) + n i)$ makes it a difference of two squares, so every $i$ cancels and the factor is real.
True or False?
The complex roots of any quadratic equation always occur as a conjugate pair.
False.
This holds only when the coefficients of the quadratic are real.
If a coefficient is itself complex the two roots need not be conjugates, and the quadratic formula has to be used to find them.
Why must a cubic equation with real coefficients always have at least one real root?
It has three roots in total, and non-real roots always arrive in conjugate pairs, so the number of non-real roots has to be even.
An even number cannot make up three on its own, so at least one root is left over and must be real.
A quartic equation with real coefficients has four roots. Complete the three possible cases:
4 real; 2 real and one $______$ pair; or $______$ complex conjugate pairs.
The completed cases are: 4 real; 2 real and one complex conjugate pair; or two complex conjugate pairs.
Notice that no case has an odd number of non-real roots, which is why 3 real and 1 non-real never appears in the list.
You are given one complex root of a cubic with real coefficients. What is the quickest route to the other two roots?
Write down the conjugate as a second root, then multiply the two factors together to get a real quadratic factor.
Dividing the cubic by that quadratic, or comparing coefficients, leaves the linear factor and so the third root.
A polynomial equation has unknown real coefficients and you are given one complex root. What do you do with the root?
Substitute it in, expand, and collect the real and imaginary parts separately.
Comparing parts then gives simultaneous equations in the unknown coefficients, and once those are known the remaining roots are found in the usual way.
On an Argand diagram, what do the horizontal and vertical axes represent?
The horizontal axis is the real axis, labelled Re, and the vertical axis is the imaginary axis, labelled Im.
The complex number $x + i y$ is plotted at the Cartesian point $(x, y)$ , so a real number lies on the horizontal axis and a purely imaginary number on the vertical axis.
Define the modulus of a complex number.
The modulus $|z|$ is the distance from the origin to the point representing $z$ on an Argand diagram.
Being a distance it is never negative, and $|z| = 0$ only when $z = 0$ .
For $z = x + i y$ complete the formula for the modulus:
$|z| = \sqrt{______+______}$
The completed formula is:
$|z| = \sqrt{x^{2} + y^{2}}$
This is Pythagoras applied to the right-angled triangle whose two shorter sides are $x$ and $y$ on an Argand diagram.
On an Argand diagram $z_{1}$ and $z_{2}$ are drawn as vectors from the origin and completed into a parallelogram. Which diagonal gives $|z_{1} + z_{2}|$ and which gives $|z_{1} - z_{2}|$ as its length?
$|z_{1} + z_{2}|$ is the diagonal starting at the origin, and $|z_{1} - z_{2}|$ is the other diagonal, the one joining the points $z_{1}$ and $z_{2}$ .
That second result is the useful one, because it says $|z_{1} - z_{2}|$ is the distance between the two points.
Define the argument of a complex number.
The argument $arg (z)$ is the anticlockwise angle, measured in radians, from the positive real axis to the line joining the origin to $z$ .
It is normally given as the principal argument, in the range $- π < arg (z) \leq π$ .
Why is the $\tan$ ratio on its own not enough to find the argument of a complex number?
The $\tan$ ratio only gives the acute angle that the line makes with the real axis, and the same value of $\tan$ occurs in two opposite quadrants.
A sketch shows which quadrant $z$ lies in, and so whether the argument is positive or negative and acute or obtuse.
True or False?
Adding two complex numbers always adds their moduli.
False.
Moduli are distances from the origin, and distances only add when the two numbers point in the same direction.
For example $3 + 4 i$ and $- 3 + 4 i$ both have modulus $5$ , yet their sum $8 i$ has modulus $8$ rather than $10$ .
Which complex numbers have a negative principal argument?
Those whose imaginary part is negative, which is the third and fourth quadrants of an Argand diagram.
Turning clockwise from the positive real axis reaches such a point sooner than turning anticlockwise, so the principal argument is recorded as negative.
Why is $arg (0)$ undefined?
The number $0$ sits at the origin, so there is no line from the origin out to it and therefore no angle to measure.
Every other complex number gives a definite direction, but $0$ gives none.
For a complex number with modulus $r$ and argument $θ$ complete the modulus-argument form:
$z = r (______+ i______)$
The completed form is:
$z = r (\cos θ + i \sin θ)$
It comes from substituting $x = r \cos θ$ and $y = r \sin θ$ into $z = x + i y$ .
Is $z = 5 (\cos \frac{π}{3} - i \sin \frac{π}{3})$ in modulus-argument form, and what is its argument?
No: modulus-argument form needs a plus sign between the two terms.
Using $\cos θ \equiv \cos (- θ)$ and $- \sin θ \equiv \sin (- θ)$ it becomes $5 (\cos (- \frac{π}{3}) + i \sin (- \frac{π}{3}))$ , so the argument is $- \frac{π}{3}$ .
How do you convert $z = 2 (\cos (- \frac{π}{3}) + i \sin (- \frac{π}{3}))$ into Cartesian form?
Work out the two trigonometric values and multiply out, using $\cos (- \frac{π}{3}) = \frac{1}{2}$ and $\sin (- \frac{π}{3}) = - \frac{\sqrt{3}}{2}$ .
This gives $z = 2 (\frac{1}{2} - \frac{\sqrt{3}}{2} i) = 1 - \sqrt{3} i$ .
Complete the two rules for multiplying complex numbers in modulus-argument form:
$|z_{1} z_{2}| =______and arg (z_{1} z_{2}) =______$
The completed rules are:
$|z_{1} z_{2}| = |z_{1}| |z_{2}| and arg (z_{1} z_{2}) = arg (z_{1}) + arg (z_{2})$
Multiplying multiplies the moduli and adds the arguments, which is what makes polar form quicker than Cartesian form for products.
True or False?
Dividing one complex number by another divides their moduli and divides their arguments.
False.
The moduli divide, but the arguments subtract.
In symbols $|\frac{z_{1}}{z_{2}}| = \frac{|z_{1}|}{|z_{2}|}$ while $arg (\frac{z_{1}}{z_{2}}) = arg (z_{1}) - arg (z_{2})$ .
Two complex numbers are multiplied and their arguments add to $\frac{7 π}{6}$ . How do you give the argument of the product as a principal argument?
Subtract $2 π$ to bring it into the range $- π < θ \leq π$ , which gives $\frac{7 π}{6} - 2 π = - \frac{5 π}{6}$ .
Adding or subtracting $2 π$ never changes which point the complex number is, only the angle used to describe it.
What does multiplying a complex number by $2 (\cos \frac{π}{2} + i \sin \frac{π}{2})$ do to it on an Argand diagram?
It doubles the number's distance from the origin and rotates it anticlockwise through $\frac{π}{2}$ .
Multiplying always scales by the modulus of the multiplier and rotates by its argument.
How does the multiplication rule give the modulus and argument of $z^{2}$ or $z^{3}$ without expanding?
Treat $z^{2}$ as $z \times z$ , so the moduli multiply and the arguments add, giving $|z^{2}| = {|z|}^{2}$ and $arg (z^{2}) = 2 arg (z)$ .
Repeating the same step for $z^{3}$ gives a modulus of ${|z|}^{3}$ and an argument of $3 arg (z)$ .
Complete the Cartesian equations of these two loci:
$Re (z) = 4 is the line______and Im (z) = - 3 is the line______$
The completed line is:
$Re (z) = 4 is the line x = 4 and Im (z) = - 3 is the line y = - 3$
The first is vertical and the second is horizontal, since fixing the real part fixes $x$ and fixing the imaginary part fixes $y$ .
Why does $|z - a| = k$ describe a circle?
Because $|z - a|$ is the distance between the points $z$ and $a$ on an Argand diagram.
Insisting that this distance is the fixed value $k$ puts $z$ on the circle of radius $k$ centred at $a$ .
The locus $|z - 8 i + 3| = 12$ is a circle. What are its centre and its radius?
Rewrite it as $|z - (- 3 + 8 i)| = 12$ so that it matches the standard form $|z - a| = k$ .
The centre is then $- 3 + 8 i$ and the radius is $12$ .
Complete the description of the locus $|z - a| = |z - b|$ :
It is the $______$ of the line joining the points $a$ and $b$ .
The completed description is: it is the perpendicular bisector of the line joining the points $a$ and $b$ .
The equation says $z$ is the same distance from $a$ as it is from $b$ , and those are exactly the points on that bisector.
Before sketching the circle $|z - (3 + 4 i)| = 10$ why should you check where the origin lies?
The sketch has to show correctly whether the circle encloses the origin.
Here the centre is a distance $5$ from the origin, by Pythagoras on $3$ and $4$ , which is less than the radius $10$ , so the origin lies inside the circle.
True or False?
The locus $arg (z - a) = α$ is drawn with an open circle at $a$ because that point is not part of the locus.
True.
At $z = a$ the expression $z - a$ is zero, and $arg (0)$ is undefined because there is no direction to measure.
The open circle is how that single excluded point is shown on the diagram.
The locus $arg (z) = \frac{π}{4}$ lies along the line $y = x$ but why is only half of that line included?
Only the points with $x > 0$ have an argument of $\frac{π}{4}$ .
The points on the other side of the origin lie in the third quadrant and have argument $- \frac{3 π}{4}$ , so they are not on the locus.
How do you write the locus $|z - (1 - 2 i)| = 3$ as a Cartesian equation?
Read off the centre $(1, - 2)$ and the radius $3$ , then put them into the standard equation of a circle.
This gives ${(x - 1)}^{2} + {(y + 2)}^{2} = 9$ .
Complete the rule for drawing the boundary of a region on an Argand diagram:
A strict inequality such as $|z| < 5$ has a $______$ boundary, while an inequality that can be equal such as $|z| \leq 5$ has a $______$ boundary.
The completed rule is: a strict inequality such as $|z| < 5$ has a dotted boundary, while one that can be equal such as $|z| \leq 5$ has a solid boundary.
A dotted line shows that the points on the boundary itself are not part of the region.
For $|z - a| < |z - b|$ which side of the perpendicular bisector do you shade?
You shade the side that contains the point $a$ .
The inequality says $z$ is closer to $a$ than to $b$ , and a useful check is that the inequality sign points like an arrow towards the letter whose side is shaded.
Complete these two half-plane regions on an Argand diagram:
$Re (z) > 3$ is the region to the $______$ of a vertical line, and $Im (z) > 4$ is the region $______$ a horizontal line.
The completed regions are: $Re (z) > 3$ is the region to the right of a vertical line, and $Im (z) > 4$ is the region above a horizontal line.
Reversing each inequality sign moves the shading to the left and to below instead.
How do the regions $|z - a| < k$ and $|z - a| > k$ differ?
$|z - a| < k$ is the region inside the circle of radius $k$ centred at $a$ , and $|z - a| > k$ is the region outside it.
In both cases the circle itself is the boundary, drawn dotted because neither inequality allows equality.
True or False?
The region $|z - a| > k$ has no greatest value of $|z|$ .
True.
The region outside a circle is unbounded, stretching away for ever in every direction, so it contains points arbitrarily far from the origin.
No greatest value of $|z|$ therefore exists, even though a least value usually does.
Describe the region given by $0 < arg (z - a) < \frac{π}{4}$ on an Argand diagram.
It is a wedge with its point at $a$ , lying between the half-lines drawn from $a$ at angles $0$ and $\frac{π}{4}$ to the positive real axis.
Both boundary half-lines are dotted here, because both inequalities are strict.
A region is the disc $|z - 5 i| \leq 2$ which does not contain the origin. How do you find the greatest and least values of $|z|$ in it?
For a circle that does not contain the origin, measure the distance from the origin to the centre, then add a radius for the greatest value and subtract a radius for the least.
Here that distance is $5$ , so the greatest $|z|$ is $7$ and the least is $3$ .
For a circular region that does not contain the origin, how do you find the greatest and least values of $arg (z)$ in it?
The extreme arguments occur where a line drawn from the origin is a tangent to the circle.
Let $θ$ be the argument of the centre and $α$ the angle between the line to the centre and a tangent, so that $\sin α = \frac{r}{d}$ for radius $r$ and centre distance $d$ .
The greatest argument is then $θ + α$ and the least is $θ - α$ .

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