# Real and Imaginary Numbers

#### Tan Tom

##### New member
Good morning,

I am working on a problem where I am finding the 4th Coefficient in a sample of 4 discrete time Fourier Series coefficients. I got the sum but now I have to solve for a_3 which consists of a real and imaginary part. Any assitance on how to solve for the a_3? Thank you.

$a_k = \{3, 1-2j, -1, ?\}$

Step 1: $(1-2j)e^{j*.5\pi*n} +a_3 e ^ {-(j*.5\pi*n)} + 3 + (-1)^{n+1}$

Step 2: $[(1-2j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2n-j \sin \frac\pi2n)]$

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Good morning,

I am working on a problem where I am finding the 4th Coefficient in a sample of 4 discrete time Fourier Series coefficients. I got the sum but now I have to solve for a_3 which consists of a real and imaginary part. Any assitance on how to solve for the a_3? Thank you.

$a_k = \{3, 1-2j, -1, ?\}$

Step 1: $(1-2j)e^{j*.5\pi*n} +a_3 e ^ {-(j*.5\pi*n)} + 3 + (-1)^{n+1}$

Step 2: $[(1-2j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2n-j \sin \frac\pi2n)]$
Hi Tan Tom, welcome to MHB! If I understand correctly, you have
$$(1-2j)e^{j\frac\pi 2 n} +a_3 e ^ {-(j\frac\pi 2n)} + 3 + (-1)^{n+1}=sum$$
for some known $sum$.

We can rewrite it as:
$$a_3 e ^ {-(j\frac\pi 2 n)}=sum-(1-2j)e^{j\frac \pi 2n} - 3 - (-1)^{n+1}\\ a_3 =\big[sum-(1-2j)e^{j\frac \pi 2n} - 3 - (-1)^{n+1}\big]e ^ {j\frac\pi 2 n}$$
Is that what you're looking for, or am I misunderstanding something?

#### HallsofIvy

##### Well-known member
MHB Math Helper
Before you can "solve" you have to have an equation! What is that supposed to be equal to? Klaas van Aarsen is assuming it is to be equal to some number he is calling "sum". Is that correct?