# Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$

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 $$\frac{9}{2}-\frac{9}{2}i$$?

## 4 Answers

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 $$1-4i+\frac{7}{2}-\frac{i}{2}=\frac{9}{2}-\frac{9i}{2}$$.

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

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

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