Simplify the expression
\[\frac{ \sqrt{-1} }{ (4-3i)-(2+5i)}\]
you need to know a symbol for a weird number \[\sqrt{-1}=i\]
you have an expression for difference of two complex numbers in the bottom first start by simplifying that.
i did earlier and got 2-8i but i didn't know what to do with it
\[\frac{ \sqrt{-1} }{ (4-3i)-(2+5i)} = \dfrac{i}{2 -8i}\] this is not in standard form yet
whats the conjugate of denominator ?
when you swithc the sign?
switch
2+8i?
yes flipping the sing of imaginary part gives you conjugate
yes we use conjugate to get rid off the i's in the denominator :
multiply top and bottom by 2+8i
i dont know what you get with i(2+8i)?
yes! \[\begin{align}\frac{ \sqrt{-1} }{ (4-3i)-(2+5i)} &= \dfrac{i}{2 -8i} \\~\\ &= \dfrac{i}{2-8i}\color{Gray}{\times \dfrac{2+8i}{2+8i}} \\~\\ &= \dfrac{i\color{Gray}{(2+8i)}}{(2-8i)\color{Gray}{(2+8i)}} \\~\\ \end{align} \]
does it flip with (i)? like -8 + 21
2i
yes, \(i=\sqrt{-1}\) so \(i^2 = -1\)
ok so \[\frac{ -8+2i }{ 68 }\]
\[\begin{align}\frac{ \sqrt{-1} }{ (4-3i)-(2+5i)} &= \dfrac{i}{2 -8i} \\~\\ &= \dfrac{i}{2-8i}\color{Gray}{\times \dfrac{2+8i}{2+8i}} \\~\\ &= \dfrac{i\color{Gray}{(2+8i)}}{(2-8i)\color{Gray}{(2+8i)}} \\~\\ &= \dfrac{2i + 8i^2}{2^2 - (8i)^2} \\~\\ &= \dfrac{2i -8}{4+64} \\~\\ \end{align} \]
Yes! just cancel the 2 top and bottomif you want.. its not so important though
\[\begin{align}\frac{ \sqrt{-1} }{ (4-3i)-(2+5i)} &= \dfrac{i}{2 -8i} \\~\\ &= \dfrac{i}{2-8i}\color{Gray}{\times \dfrac{2+8i}{2+8i}} \\~\\ &= \dfrac{i\color{Gray}{(2+8i)}}{(2-8i)\color{Gray}{(2+8i)}} \\~\\ &= \dfrac{2i + 8i^2}{2^2 - (8i)^2} \\~\\ &= \dfrac{2i -8}{4+64} \\~\\ &= \dfrac{-4+i}{34} \\~\\ &=-\dfrac{2}{17} + \dfrac{1}{34}i \end{align} \]
is it \[\frac{ -4+i }{ 34 }\]
i forgot to write the answer choices lol 15 - 8i / 49 -4 + i / 34 15 - 8i / 79 -4 + i / 4
looks good!
so B is right?
-4 + 1 / 34
(-4 + \(\color{red}{i}\)) / 34
yes
okkkk thank you so much. that's all i was stuck on, you're a great help! :)
and just turned this in and got a 100%, thanks again
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