Cosine Complex 9468E9

1. **State the problem:** We need to analyze and draw the trigonometric graphs of the function $$y = \cos\left(x + (period)bi\right)$$ where $b$ is a real number and $i$ is the imaginary unit. 2. **Recall the formula and properties:** The cosine function for a complex argument $z = x + yi$ can be expressed using Euler's formula: $$\cos(z) = \cos(x + yi) = \cos x \cosh y - i \sin x \sinh y$$ where $\cosh y$ and $\sinh y$ are the hyperbolic cosine and sine functions. 3. **Apply to our function:** Here, the argument is $x + (period)bi$. Assuming $(period)$ is a constant real number representing the period of cosine (usually $2\pi$), the argument is complex: $$z = x + (period) b i$$ 4. **Rewrite the function:** Using the formula, $$y = \cos(x + (period) b i) = \cos x \cosh((period) b) - i \sin x \sinh((period) b)$$ 5. **Interpretation:** The function is complex-valued. The real part is: $$\text{Re}(y) = \cos x \cosh((period) b)$$ and the imaginary part is: $$\text{Im}(y) = - \sin x \sinh((period) b)$$ 6. **Graphing:** To graph this, plot separately: - The real part: $y_1 = \cos x \cosh((period) b)$ - The imaginary part: $y_2 = - \sin x \sinh((period) b)$ 7. **Summary:** The graphs are scaled versions of $\cos x$ and $\sin x$ by hyperbolic functions of $(period) b$. **Final answer:** The trigonometric graphs of $y = \cos(x + (period)bi)$ are given by the real and imaginary parts: $$\text{Re}(y) = \cos x \cosh((period) b), \quad \text{Im}(y) = - \sin x \sinh((period) b)$$