2.4 Complex numbers  (Page 4/8)

Explain how to add complex numbers.

What is the basic principle in multiplication of complex numbers?

Give an example to show that the product of two imaginary numbers is not always imaginary.

What is a characteristic of the plot of a real number in the complex plane?

If $y=x^2+x-4,$ evaluate $y$ given $x=2i.$

If $y=x^3-2,$ evaluate $y$ given $x=i.$

If $y=x^2+3x+5,$ evaluate $y$ given $x=2+i.$

If $y=2x^2+x-3,$ evaluate $y$ given $x=2-3i.$

If $y=\frac{x+1}{2-x},$ evaluate $y$ given $x=5i.$

If $y=\frac{1+2x}{x+3},$ evaluate $y$ given $x=4i.$

$1-2i$

$\mathrm{-2}+3i$

$i$

$\mathrm{-3}-4i$

$\left(3+2i\right)+(5-3i)$

$\left(\mathrm{-2}-4i\right)+\left(1+6i\right)$

$\left(\mathrm{-5}+3i\right)-(6-i)$

$\left(2-3i\right)-(3+2i)$

$(\mathrm{-4}+4i)-(\mathrm{-6}+9i)$

$\left(2+3i\right)(4i)$

$\left(5-2i\right)(3i)$

$\left(6-2i\right)(5)$

$\left(\mathrm{-2}+4i\right)\left(8\right)$

$\left(2+3i\right)(4-i)$

$\left(\mathrm{-1}+2i\right)(\mathrm{-2}+3i)$

$\left(4-2i\right)(4+2i)$

$\left(3+4i\right)\left(3-4i\right)$

$\frac{3+4i}{2}$

$\frac{6-2i}{3}$

$\frac{-5+3i}{2i}$

$\frac{6+4i}{i}$

$\frac{2-3i}{4+3i}$

$\frac{3+4i}{2-i}$

$\frac{2+3i}{2-3i}$

$\sqrt{\mathrm{-9}}+3\sqrt{\mathrm{-16}}$

$-\sqrt{\mathrm{-4}}-4\sqrt{\mathrm{-25}}$

$\frac{2+\sqrt{-12}}{2}$

$\frac{4+\sqrt{-20}}{2}$

$i^8$

$i^{15}$

$i^{22}$

Evaluate ${(1+i)}^k$ for $k=4,8,\text{and}12.$ Predict the value if $k=16.$

Evaluate ${(1-i)}^k$ for $k=2,6,\text{and}10.$ Predict the value if $k=14.$

Evaluate ${(\text{l}+i)}^k-{(\text{l}-i)}^k$ for $k=4,8,\text{and}12.$ Predict the value for $k=16.$

Show that a solution of $x^6+1=0$ is $\frac{\sqrt{3}}{2}+\frac{1}{2}i.$

Show that a solution of $x^8\mathrm{-1}=0$ is $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i.$

$\frac{1}{i}+\frac{4}{i^3}$

$\frac{1}{i^{11}}-\frac{1}{i^{21}}$

$i^7\left(1+i^2\right)$

$i^{\mathrm{-3}}+5i^7$

$\frac{\left(2+i\right)\left(4-2i\right)}{(1+i)}$

$\frac{\left(1+3i\right)\left(2-4i\right)}{(1+2i)}$

$\frac{{\left(3+i\right)}^2}{{\left(1+2i\right)}^2}$

$\frac{3+2i}{2+i}+\left(4+3i\right)$

$\frac{4+i}{i}+\frac{3-4i}{1-i}$

$\frac{3+2i}{1+2i}-\frac{2-3i}{3+i}$