Example

Simplify [latex]\sqrt{-9}[/latex].

We can separate [latex]\sqrt{-9}[/latex] as [latex]\sqrt{9}\sqrt{-1}[/latex]. We can take the square root of 9, and write the square root of [latex]-1[/latex] as i.

[latex]\sqrt{-9}=\sqrt{9}\sqrt{-1}=3i[/latex]

Example

Plot the number [latex]3-4i[/latex] on the complex plane.

The real part of this number is 3, and the imaginary part is [latex]-4[/latex]. To plot this, we draw a point 3 units to the right of the origin in the horizontal direction and 4 units down in the vertical direction.

Example

Add [latex]3-4i[/latex] and [latex]2+5i[/latex].

Adding [latex](3-4i)+(2+5i)[/latex], we add the real parts and the imaginary parts.

[latex]3+2-4i+5i[/latex]

[latex]5+i[/latex]

Example

Visualize the addition [latex]3-4i[/latex] and [latex]-1+5i[/latex].

The initial point is [latex]3-4i[/latex]. When we add [latex]-1+3i[/latex], we add [latex]-1[/latex] to the real part, moving the point 1 units to the left, and we add 5 to the imaginary part, moving the point 5 units vertically. This shifts the point [latex]3-4i[/latex] to [latex]2+1i[/latex].

Example

Multiply: [latex]4\left(2+5i\right)[/latex]

To multiply the complex number by a real number, we simply distribute as we would when multiplying polynomials.

Distribute and simplify.

[latex]4(2+5i)\\\,\,\,= 4\cdot2+4\cdot5i\\\,\,\,=8+20i[/latex]

Example

Multiply: [latex](2+5i)(4+i)[/latex].

[latex]\begin{array}{l}\left(2+5i\right)\left(4+i\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Expand.}\\=8+20i+2i+5i^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Since }i=\sqrt{-1},i^{2}=-1\\=8+20i+2i+5\left(-1\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Simplify.}\\=3+22i\end{array}[/latex]

Example

Visualize the product [latex]2(1+2i)[/latex].

Multiplying we’d get

[latex]\begin{align}&2\cdot1+2\cdot2i\\&=2+4i\\\end{align}[/latex]

Notice both the real and imaginary parts have been scaled by 2. Visually, this will stretch the point outwards, away from the origin.

Example

Visualize the product [latex]i\left(l+2i\right)[/latex].

Multiplying, we’d get

[latex]\begin{align}&i\cdot1+i\cdot2i\\&=i+2{{i}^{2}}\\&=i+2(-1)\\&=-2+i\\\end{align}[/latex]

In this case, the distance from the origin has not changed, but the point has been rotated about the origin, 90° counter-clockwise.

Example

Visualize the result of multiplying [latex]1+2i[/latex] by [latex]1+i[/latex]. Then show the result of multiplying by [latex]1+i[/latex] again.

Multiplying [latex]1+2i[/latex] by [latex]1+i[/latex],

[latex]-4+2i[/latex]

[latex]\begin{align}&(1+2i)(1+i)\\&=1+i+2i+2{{i}^{2}}\\&=1+3i+2(-1)\\&=-1+3i\\\end{align}[/latex]

Multiplying by [latex]1+i[/latex] again,

[latex]\begin{align}&(-1+3i)(1+i)\\&=-1-i+3i+3{{i}^{2}}\\&=-1+2i+3(-1)\\&=-4+2i\\\end{align}[/latex]

If we multiplied by [latex]1+i[/latex] again, we’d get [latex]–6–2i[/latex]. Plotting these numbers in the complex plane, you may notice that each point gets both further from the origin, and rotates counterclockwise, in this case by 45°.

Example

Given the recursive relationship [latex]{{z}_{n+1}}={{z}_{n}}+2,\quad{{z}_{0}}=4[/latex], generate several terms of the recursive sequence.

We are given the starting value, [latex]{{z}_{0}}=4[/latex]. The recursive formula holds for any value of n, so if [latex]n = 0[/latex], then [latex]{{z}_{n+1}}={{z}_{n}}+2[/latex] would tell us [latex]{{z}_{0+1}}={{z}_{0}}+2[/latex], or more simply, [latex]{{z}_{1}}={{z}_{0}}+2[/latex].

Notice this defines [latex]{{z}_{1}}[/latex] in terms of the known [latex]{{z}_{0}}[/latex], so we can compute the value:

[latex]{{z}_{1}}={{z}_{0}}+2=4+2=6[/latex].

Now letting [latex]n = 1[/latex], the formula tells us [latex]{{z}_{1+1}}={{z}_{1}}+2[/latex], or [latex]{{z}_{2}}={{z}_{1}}+2[/latex]. Again, the formula gives the next value in the sequence in terms of the previous value.

[latex]{{z}_{2}}={{z}_{1}}+2=6+2=8[/latex]

Continuing,

[latex]{{z}_{3}}={{z}_{2}}+2=8+2=10\\{{z}_{4}}={{z}_{3}}+2=10+2=12[/latex]

Example

Given the recursive relationship [latex]{{z}_{n+1}}={{z}_{n}}\cdot{i}+(1-i),\quad{{z}_{0}}=4[/latex], generate several terms of the recursive sequence.

We are given [latex]{{z}_{0}}=4[/latex]. Using the recursive formula:

[latex]{{z}_{1}}={{z}_{0}}\cdot{i}+(1-i)=4\cdot{i}+(1-i)=1+3i\\{{z}_{2}}={{z}_{1}}\cdot{i}+(1-i)=(1+3i)\cdot{i}+(1-i)=i+3{{i}^{2}}+(1-i)=i-3+(1-i)=-2\\{{z}_{3}}={{z}_{2}}\cdot{i}+(1-i)=(-2)\cdot{i}+(1-i)=-2i+(1-i)=1-3i\\{{z}_{4}}={{z}_{3}}\cdot{i}+(1-i)=(1-3i)\cdot{i}+(1-i)=i-3{{i}^{2}}+(1-i)=i+3+(1-i)=4\\{{z}_{5}}={{z}_{4}}\cdot{i}+(1-i)=4\cdot{i}+(1-i)=1+3i[/latex]

Notice this sequence is exhibiting an interesting pattern—it began to repeat itself.

Example

Determine if [latex]c=1+i[/latex] is part of the Mandelbrot set.

We start with [latex]{{z}_{0}}=0[/latex]. We continue, omitting some detail of the calculations

[latex]{{z}_{1}}={{z}_{0}}^{2}+1+i=0+1+i=1+i\\{{z}_{2}}={{z}_{1}}^{2}+1+i={{(1+i)}^{2}}+1+i=1+3i\\{{z}_{3}}={{z}_{2}}^{2}+1+i={{(1+3i)}^{2}}+1+i=-7+7i\\{{z}_{4}}={{z}_{3}}^{2}+1+i={{(-7+7i)}^{2}}+1+i=1-97i[/latex]

We can already see that these values are getting quite large. It does not appear that [latex]c=1+i[/latex] is part of the Mandelbrot set.

Example

Determine if [latex]c=0.5i[/latex] is part of the Mandelbrot set.

We start with [latex]{{z}_{0}}=0[/latex]. We continue, omitting some detail of the calculations

[latex]{{z}_{1}}={{z}_{0}}^{2}+0.5i=0+0.5i=0.5i\\{{z}_{2}}={{z}_{1}}^{2}+0.5i={{(0.5i)}^{2}}+0.5i=-0.25+0.5i\\{{z}_{3}}={{z}_{2}}^{2}+0.5i={{(-0.25+0.5i)}^{2}}+0.5i=-0.1875+0.25i\\{{z}_{4}}={{z}_{3}}^{2}+0.5i={{(-0.1875+0.25i)}^{2}}+0.5i=-0.02734+0.40625i[/latex]

While not definitive with this few iterations, it does appear that this value is remaining small, suggesting that 0.5i is part of the Mandelbrot set.