1. The number is already in the form [latex]a+bi[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]2-i\sqrt{5}[/latex].
2. We can rewrite this number in the form [latex]a+bi[/latex] as [latex]0-\frac{1}{2}i[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]0+\frac{1}{2}i[/latex]. This can be written simply as [latex]\frac{1}{2}i[/latex].

Analysis of the Solution

Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by [latex]i[/latex].

We begin by writing the problem as a fraction.

[latex]\dfrac{\left(2+5i\right)}{\left(4-i\right)}[/latex]

Then we multiply the numerator and denominator by the complex conjugate of the denominator.

[latex]\dfrac{\left(2+5i\right)}{\left(4-i\right)}\cdot \dfrac{\left(4+i\right)}{\left(4+i\right)}[/latex]

To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).

[latex]\begin{align}\dfrac{\left(2+5i\right)}{\left(4-i\right)}\cdot \dfrac{\left(4+i\right)}{\left(4+i\right)}&=\dfrac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\[2mm] &=\dfrac{8+2i+20i+5\left(-1\right)}{16+4i - 4i-\left(-1\right)} && \text{Because } {i}^{2}=-1 \\[2mm] &=\frac{3+22i}{17} \\[2mm] &=\dfrac{3}{17}+\frac{22}{17}i && \text{Separate real and imaginary parts}.\end{align}[/latex]

Note that this expresses the quotient in standard form.

Substitute [latex]x=3+i[/latex] into the function [latex]f\left(x\right)={x}^{2}-5x+2[/latex] and simplify.

[latex]\begin{align}f(3+i)&=(3+i)^2-5(3+i)+2&&\text{Substitute } 3+i \text{ for }x\\[2mm]&=(3+6i+i^2)-(15+5i)+2&&\text{Multiply}\\[2mm]&=9+6i+(-1)-15-5i+2&&\text{Substitute }-1\text{ for }i^2 \\[2mm]&=-5+i&&\text{Combine like terms}\end{align}[/latex]

Analysis of the Solution

We write [latex]f\left(3+i\right)=-5+i[/latex]. Notice that the input is [latex]3+i[/latex] and the output is [latex]-5+i[/latex].

Substitute [latex]x=10i[/latex] and simplify.

[latex]\begin{align}&\dfrac{2+10i}{10i+3} && \text{Substitute }10i\text{ for }x\\[2mm] &\dfrac{2+10i}{3+10i} && \text{Rewrite the denominator in standard form}\\[2mm] &\dfrac{2+10i}{3+10i}\cdot \dfrac{3 - 10i}{3 - 10i} && \text{Multiply the numerator and denominator by the complex conjugate of the denominator}\\[2mm] &\dfrac{6 - 20i+30i - 100{i}^{2}}{9 - 30i+30i - 100{i}^{2}} && \text{Multiply using the distributive property or the FOIL method} \\[2mm] &\dfrac{6 - 20i+30i - 100\left(-1\right)}{9 - 30i+30i - 100\left(-1\right)} && \text{Substitute }-1\text{ for } {i}^{2} \\[2mm] &\dfrac{106+10i}{109} && \text{Simplify} \\[2mm] &\dfrac{106}{109}+\dfrac{10}{109}i && \text{Separate the real and imaginary parts} \end{align}[/latex]

Since [latex]{i}^{4}=1[/latex], we can simplify the problem by factoring out as many factors of [latex]{i}^{4}[/latex] as possible. To do so, first determine how many times [latex]4[/latex] goes into [latex]35[/latex]: [latex]35=4\cdot 8+3[/latex].

[latex]{i}^{35}={i}^{4\cdot 8+3}={i}^{4\cdot 8}\cdot {i}^{3}={\left({i}^{4}\right)}^{8}\cdot {i}^{3}={1}^{8}\cdot {i}^{3}={i}^{3}=-i[/latex]