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

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

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

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

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.