Porisma

Worked solution

Plot six complex operations and explain their geometry

Given z1=1+3iz_1=1+3i and z2=4+3iz_2=4+3i, mark and label clearly on Argand diagrams the points representing:

(i) 3z13z_1 and −2z2-2z_2;

(ii) z1z2z_1z_2;

(iii) −z2-z_2;

(iv) z1+z2z_1+z_2;

(v) z1−z2z_1-z_2; and

(vi) iz1iz_1, i2z1i^2z_1 and −iz1-iz_1.

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Equal-unit Argand grid showing only the given input points z1=(1,3) and z2=(4,3). Calculate and label the requested outputs; six separate solution plots retain the source constructions for parts(i)–(vi).
Equal-unit Argand grid showing only the given input points z1=(1,3) and z2=(4,3). Calculate and label the requested outputs; six separate solution plots retain the source constructions for parts(i)–(vi).
Equal-unit Argand grid showing only the given input points z1=(1,3) and z2=(4,3). Calculate and label the requested outputs; six separate solution plots retain the source constructions for parts(i)–(vi).

CJC · Lecture · Example 2

(i) Scale and label both points

Answer

3z1=3+9i3z_1=3+9i at (3,9)(3,9); −2z2=−8−6i-2z_2=-8-6i at (−8,−6)(-8,-6).

The source calculations are 3z1=3(1+3i)=3+9i3z_1=3(1+3i)=3+9i and −2z2=−2(4+3i)=−8−6i-2z_2=-2(4+3i)=-8-6i. Plot real parts horizontally and imaginary parts vertically. Positive scaling by 3 keeps the direction; negative scaling by −2 reverses the direction and doubles the distance from O.

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Equal-unit Argand grid: vectors from O to 3z1=(3,9) and −2z2=(−8,−6), each with its point and complex-number label.
Equal-unit Argand grid: vectors from O to 3z1=(3,9) and −2z2=(−8,−6), each with its point and complex-number label.
Equal-unit Argand grid: vectors from O to 3z1=(3,9) and −2z2=(−8,−6), each with its point and complex-number label.
How did this part go?

(ii) Multiply, then plot the product

Answer

z1z2=−5+15iz_1z_2=-5+15i at (−5,15)(-5,15).

Expand every term before using i2=−1i^2=-1:

z1z2=(1+3i)(4+3i)=4+12i+3i+9i2=4+15i+9(−1)=−5+15i.\begin{aligned}z_1z_2&=(1+3i)(4+3i)\\&=4+12i+3i+9i^2\\&=4+15i+9(-1)\\&=-5+15i.\end{aligned}

The source’s vector remark distinguishes this from the scalar product of the corresponding coordinate vectors: (1,3)⋅(4,3)=4+9=13(1,3)\cdot(4,3)=4+9=13. The complex product is −5+15i-5+15i, represented by (−5,15)(-5,15). These are different operations and different outputs; multiplying complex numbers is not a vector dot product.

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Equal-unit Argand grid retains original z1=(1,3) and z2=(4,3) vectors and the product vector to z1z2=(−5,15).
Equal-unit Argand grid retains original z1=(1,3) and z2=(4,3) vectors and the product vector to z1z2=(−5,15).
Equal-unit Argand grid retains original z1=(1,3) and z2=(4,3) vectors and the product vector to z1z2=(−5,15).
How did this part go?

(iii) Negate and interpret the point

Answer

−z2=−4−3i-z_2=-4-3i at (−4,−3)(-4,-3).

The source gives −z2=−(4+3i)=−4−3i-z_2=-(4+3i)=-4-3i. Its coordinates are the negatives of those of z2z_2, so the point is a 180° rotation of z2z_2 about the origin O. Negation retains the distance from O and reverses the direction.

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Equal-unit Argand grid: z2=(4,3) and −z2=(−4,−3) lie opposite each other on the same line through O, showing a 180-degree rotation.
Equal-unit Argand grid: z2=(4,3) and −z2=(−4,−3) lie opposite each other on the same line through O, showing a 180-degree rotation.
Equal-unit Argand grid: z2=(4,3) and −z2=(−4,−3) lie opposite each other on the same line through O, showing a 180-degree rotation.
How did this part go?

(iv) Add by the parallelogram law

Answer

z1+z2=5+6iz_1+z_2=5+6i at (5,6)(5,6).

The source calculation is z1+z2=(1+3i)+(4+3i)=5+6iz_1+z_2=(1+3i)+(4+3i)=5+6i. The two original vectors form adjacent sides of a parallelogram; the diagonal from O represents their sum. Equivalently, add z2z_2 by translating the point representing z1z_1 by (4,3)(4,3), or add z1z_1 by translating z2z_2 by (1,3)(1,3). The source describes addition as translation from one point to another.

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Equal-unit Argand grid: O, z1=(1,3), z2=(4,3), and sum=(5,6) form a parallelogram. Translated side arrows join each original point to the sum; the diagonal is the sum vector.
Equal-unit Argand grid: O, z1=(1,3), z2=(4,3), and sum=(5,6) form a parallelogram. Translated side arrows join each original point to the sum; the diagonal is the sum vector.
Equal-unit Argand grid: O, z1=(1,3), z2=(4,3), and sum=(5,6) form a parallelogram. Translated side arrows join each original point to the sum; the diagonal is the sum vector.
How did this part go?

(v) Subtract by adding the negative

Answer

z1−z2=−3z_1-z_2=-3 at (−3,0)(-3,0).

The source gives z1−z2=(1+3i)−(4+3i)=−3z_1-z_2=(1+3i)-(4+3i)=-3. Subtraction is addition of −z2-z_2: translate the point z1z_1 by (−4,−3)(-4,-3) to reach (−3,0)(-3,0). The source also shows the parallelogram with sides z1z_1 and −z2z_2, giving their sum as the diagonal. The imaginary coefficient is zero, so this result lies on the real axis.

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Equal-unit Argand grid retains z1=(1,3), z2=(4,3), −z2=(−4,−3) and difference=(−3,0), with the subtraction parallelogram and a diagonal from O to the difference.
Equal-unit Argand grid retains z1=(1,3), z2=(4,3), −z2=(−4,−3) and difference=(−3,0), with the subtraction parallelogram and a diagonal from O to the difference.
Equal-unit Argand grid retains z1=(1,3), z2=(4,3), −z2=(−4,−3) and difference=(−3,0), with the subtraction parallelogram and a diagonal from O to the difference.
How did this part go?

(vi) Label the three rotations

Answer

iz1=−3+iiz_1=-3+i, i2z1=−1−3ii^2z_1=-1-3i, and −iz1=3−i-iz_1=3-i.

Follow all three printed calculations:

iz1=i(1+3i)=−3+i,i2z1=i(−3+i)=−1−3i,−iz1=i3z1=i(−1−3i)=3−i.\begin{aligned}iz_1&=i(1+3i)=-3+i,\\i^2z_1&=i(-3+i)=-1-3i,\\-iz_1&=i^3z_1=i(-1-3i)=3-i.\end{aligned}

Starting from z1z_1 at (1,3)(1,3), the three points are (−3,1)(-3,1), (−1,−3)(-1,-3) and (3,−1)(3,-1). Multiplication by i rotates 90° anticlockwise, multiplication by i2=−1i^2=-1 rotates 180°, and multiplication by −i rotates 90° clockwise, all about O. Each has the same modulus as z1z_1. The source recommends drawing every Argand diagram to scale to avoid a distorted view; the plots here use equal x- and y-unit lengths.

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Equal-unit Argand grid: z1=(1,3), iz1=(−3,1), i²z1=(−1,−3), −iz1=(3,−1). All vectors start at O and have equal length; right-angle markers show the quarter-turn relationships.
Equal-unit Argand grid: z1=(1,3), iz1=(−3,1), i²z1=(−1,−3), −iz1=(3,−1). All vectors start at O and have equal length; right-angle markers show the quarter-turn relationships.
Equal-unit Argand grid: z1=(1,3), iz1=(−3,1), i²z1=(−1,−3), −iz1=(3,−1). All vectors start at O and have equal length; right-angle markers show the quarter-turn relationships.
How did this part go?

Solution notes

All six printed asks and every source calculation/geometric interpretation are retained. Equal-unit grids preserve the source recommendation to draw to scale. The source Student copy already prints calculations and several constructions; the Teacher adds the remaining plotted answer points.