9.3 Homework: complex numbers

$(3+7i)–(4+7i)=$

$(5–3i)+(5–3i)=$

$2(5–3i)=$

$(5–3i)(2+0i)=$

What is the complex conjugate of $(5–3i)$ ?

What do you get when you multiply $(5–3i)$ by its complex conjugate?

What is the complex conjugate of 7?

What do you get when you multiply 7 by its complex conjugate?

What is the complex conjugate of $2i$ ?

What do you get when you multiply $2i$ by its complex conjugate?

What is the complex conjugate of $(a+bi)$ ?

What do you get when you multiply $(a+bi)$ by its complex conjugate?

I’m thinking of a complex number $z$ . When I multiply it by its complex conjugate (designated as $z*$ ) the answer is 25.

• A

  What might $z$ be?

• B

  Test it, and make sure it works—that is, that $\left(z\right)\left(z*\right)=25$ !

I’m thinking of a different complex number $z$ . When I multiply it by its complex conjugate, the answer is $3+2i$ .

• A

  What might $z$ be?

• B

  Test it, and make sure it works—that is, that $\left(z\right)\left(z*\right)=3+2i$ !

Solve for $x$ and $y$ : $x^2+2x^{2}i+4y+40yi=7–2i$

Finally, a bit more exercise with rational expressions. We’re going to take one problem and solve it two different ways. The problem is $\frac{3}{2+i}-\frac{\mathrm{7i}}{3+\mathrm{4i}}$ . The final answer, of course, must be in the form $a+bi$ .

• A

  Here is one way to solve it: the common denominator is $(2+i)(3+4i)$ . Put both fractions over the common denominator and combine them. Then, take the resulting fraction, and simplify it into $a+bi$ form.

• B

  Here is a completely different way to solve the same problem. Take the two fractions we are subtracting and simplify them both into $a+bi$ form, and then subtract.

• C

  Did you get the same answer? (If not, something went wrong…) Which way was easier?