Question

30+5i / 30-5i

Answer 1

multiply the numerator and denominator by the complex conjugate of the denominator. that will turn the denominator into a real number.

Answer 2

does it make sense or do you need more explanation?

Answer 3

900+10i+25i(squared) over 900+25i(squared)?

Answer 4

\[\frac{30+5i}{30-5i}=\frac{30+5i}{30-5i}*\frac{30+5i}{30+5i}=\frac{(30+5i)(30+5i)}{30^2+5^2}\]

Answer 5

remember:\[(a-bi)(a+bi)=a^2-b^2i^2=a^2-b^2(-1)=a^2+b^2\]

Answer 6

OK, this whole (i) thing is confusing to me. I always forget to use the one when I multiply. thank you

Answer 7

yw - you just need practice - I'm sure you'll get the hang of these soon :)

Answer 8

30 + ((5 * i) / 30) - (5 * i) = 30 - 4.833333

Answer 9

Squareroot of -15 times Squareroot of -15

Answer 10

anymore?

Answer 11

15i ?

Answer 12

yes

Answer 13

if im wrong idk what i did wrong

Answer 14

\[\begin{align}
\frac{30+5i}{30-5i}&=\frac{30+5i}{30-5i}*\frac{30+5i}{30+5i}=\frac{(30+5i)(30+5i)}{30^2+5^2}\\
&=\frac{30^2+2*30*5i+(5i)^2}{900+25}\\
&=\frac{900+300i+(25i^2)}{900+25}\\
&=\frac{900+300i+(25*(-1))}{925}\\
&=\frac{900+300i-25}{925}\\
&=\frac{875+300i}{925}\\
&=\frac{35+12i}{37}\\
\end{align}\]