Evaluate the following complex numbers, expressiong each answer in the form a + bi
for a, b real.
(a) 1/(2 − 3i)

Question

Evaluate the following complex numbers, expressiong each answer in the form a + bi
for a, b real.
(a) 1/(2 − 3i)

myininaya · Answer

Multiply top and bottom by 2+3i

myininaya · Answer

1/(c-di) 
Always do times (c+di) on top and bottom

myininaya · Answer

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

anonymous · Answer

ok thank you! and then how would i express that?
or evaluate?! are you able to help with 
e^(1+iπ)

myininaya · Answer

So top is easy you get 2+3i
on bottom since we are multiplying complex conjugates we get
$$(a-bi)(a+bi)=a^2+abi-abi-b^2i^2=a^2+0-b^2(-1)=a^2+b^2$$

anonymous · Answer

ok, so we replace with -1, got it!

myininaya · Answer

Are you saying you want to put it in polar form also?

myininaya · Answer

So can you tell me what we get for the bottom?

anonymous · Answer

Evaluate the following complex numbers, expressiong each answer in the form a + bi
for a, b real. that's just what its looking for, so i'm not sure if that means polar as well. for the bottom or the top? i thought you just did the bottom ahah.

myininaya · Answer

Whatever ever we do to the bottom we must do to the top 
(why were you talking about e^(1+ipi) then?-i'm confused)

anonymous · Answer

oh, i have to evaluate that one as well, but im really not sure how to go about it, and was wondering if you could help me with it.

ok, so what we did with the bottom, so what the final answer for 1 / 2-3i
would be : 2+3sqrt(-1) / 15  ?

myininaya · Answer

$$\frac{1}{2-3i} \cdot \frac{2+3i}{2+3i}=\frac{2+3i}{(2-3i)(2+3i)}$$
what does the bottom simplify to?
You said 15?
4+9=13 not 15
so we have
$$\frac{2+3i}{13}$$

myininaya · Answer

Separate the fraction and you got it!

anonymous · Answer

ok so i have to separate the fraction? would that be expressed in terms of a, b real if it still contains the i in there?

myininaya · Answer

(a+b)/c=a/c+b/c
Did you do that?

myininaya · Answer

now if you have (a+bi)/c you do the samething
a/c+bi/c

anonymous · Answer

oh ok, and once we split the fractions up its in real condition?
also, are you able to help me out with e^(1+ipi)

myininaya · Answer

I don't know what you want me to do with that honestly

myininaya · Answer

Are you wanting me to help you put it in a+bi form?

anonymous · Answer

i think the point of this exercise is to get it out of a+bi form and into a, b form as real numbers