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I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$

The provided solution is: $\frac{9}{2}-\frac{9}{2}i$

I arrived at a divide by zero error which must be incorrect. My working:

$\frac{4+i}{i}$, complex conjugate is $-i$ so:

$\frac{-i(4+i)}{-i*i}$

= $\frac{-4i+i^2}{i^2}$

= $\frac{-4i--1}{-1}$

= $-4i+1$

Then the next part:

$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:

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

= $\frac{3-4i+3i-4i^2}{1-i^2}$

= $\frac{7-i}{0}$ # 1 + -1 = 0

How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?

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You've missed a minus sign there:

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

The second one is

$\frac{3-4i}{1-i}=\frac{(3-4i)(1+i)}{(1-i)(1+i)}=\frac{7-i}{2}$

Hence the sum is $1-4i+\frac{7}{2}-\frac{i}{2}=\frac{9}{2}-\frac{9i}{2}$.

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Here is your error $1-i^2=1-(-1)=2\ne 0$

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Hint: It is $$\frac{4+i}{i}+\frac{3-4i}{1-i}=\frac{(4+i)(1-i)+(3-4i)i}{i(1-i)}$$

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You have made a mistake in the second last step of your simplification. $$1 - i^2 = 1 - (-1)=1 + 1 = 2$$ Then, $$-4i + 1 + \frac{7 - i}{2} = 1 + \frac{7}{2} + i(-4 - \frac{1}{2}) = \frac{9}{2} - \frac{9}{2}i$$

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