Let's find the **Maclaurin series** (Taylor series at \( z = 0 \)) for each function step-by-step: --- ### (a) \(\frac{1}{z+1}\) #### **Step-by-step:** Recall the geometric series: \[ \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n, \quad |x| < 1 \] Here, substitute \(x = -z\): \[ \frac{1}{z+1} = \frac{1}{1 + z} = \sum_{n=0}^{\infty} (-z)^n = \sum_{n=0}^{\infty} (-1)^n z^n, \quad |z| < 1 \] #### **Final answer:** \[ \boxed{ \frac{1}{z+1} = \sum_{n=0}^{\infty} (-1)^n z^n } \] --- ### (b) \(\frac{1}{\sqrt{1-z}}\) #### **Step-by-step:** Recall the binomial series for any real \(\alpha\): \[ (1 + x)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n, \quad |x| < 1 \] Here, \(\alpha = -\frac{1}{2}\), \(x = -z\): \[ (1-z)^{-\frac{1}{2}} = \sum_{n=0}^\infty \binom{-1/2}{n}(-z)^n \] Where \[ \binom{-1/2}{n} = \frac{(-1/2)(-1/2-1)\cdots(-1/2-n+1)}{n!} \] #### **First few terms:** - \(n=0\): \(1\) - \(n=1\): \(\left(-\frac{1}{2}\right)(-z) = \frac{1}{2}z\) - \(n=2\): \(\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\frac{(-z)^2}{2!} = \frac{3}{8}z^2\) - \(n=3\): \(\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\frac{(-z)^3}{3!} = \frac{5}{16}z^3\) So, \[ \frac{1}{\sqrt{1-z}} = 1 + \frac{1}{2}z + \frac{3}{8}z^2 + \frac{5}{16}z^3 + \cdots \] #### **Final answer:** \[ \boxed{ \frac{1}{\sqrt{1-z}} = \sum_{n=0}^{\infty} \binom{-1/2}{n} (-z)^n } \] --- ### (c) \(\frac{e^{jz} + e^{-jz}}{2}\) #### **Step-by-step:** This is the definition of the **cosine** function: \[ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} \] The Maclaurin series for \(\cos(z)\) is: \[ \cos(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{2n} \] If your \(j\) is the imaginary unit (\(j^2 = -1\)), then substitute \(z\) with \(jz\): \[ \cos(jz) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(jz)^{2n} = \sum_{n=0}^\infty \frac{(-1)^n j^{2n}}{(2n)!}z^{2n} \] But \(j^{2n} = (-1)^n\), so: \[ \cos(jz) = \sum_{n=0}^\infty \frac{(-1)^n \cdot (-1)^n}{(2n)!}z^{2n} = \sum_{n=0}^\infty \frac{1}{(2n)!}z^{2n} \] #### **Final answer:** \[ \boxed{ \frac{e^{jz} + e^{-jz}}{2} = \sum_{n=0}^\infty \frac{1}{(2n)!}z^{2n} } \] --- ### (d) \(\arccos(z)\) #### **Step-by-step:** The Maclaurin series for \(\arccos(z)\): \[ \arccos(z) = \frac{\pi}{2} - \sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}z^{2n+1} \] #### **Final answer:** \[ \boxed{ \arccos(z) = \frac{\pi}{2} - \sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}z^{2n+1} } \] --- ## **Summary Table** | Function | Maclaurin Series | |----------------------------------|----------------------------------------------------------------------------------------------| | \(\frac{1}{z+1}\) | \(\sum_{n=0}^{\infty} (-1)^n z^n\) | | \(\frac{1}{\sqrt{1-z}}\) | \(\sum_{n=0}^\infty \binom{-1/2}{n}(-z)^n\) | | \(\frac{e^{jz} + e^{-jz}}{2}\) | \(\sum_{n=0}^\infty \frac{1}{(2n)!}z^{2n}\) | | \(\arccos(z)\) | \(\frac{\pi}{2} - \sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}z^{2n+1}\) | If you need further explanations or more terms, let me know!

1. \(\displaystyle \frac{1}{z+1} = \sum_{n=0}^{\infty} (-1)^n z^n\) 2. \(\displaystyle \frac{1}{\sqrt{1 - z}} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} (-z)^n\) 3. \(\displaystyle \frac{e^{jz} + e^{-jz}}{2} = \sum_{n=0}^\infty \frac{z^{2n}}{(2n)!}\) 4. \(\displaystyle \arccos(z) = \frac{\pi}{2} - \sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1)} z^{2n+1}\)