# Math Help - complex numbers

1. ## complex numbers

Express in the form a + bi, where a and b are real

(a) sin(3pi/2 - i)

(b) cosh(1 + i)

have tried using addition formula on (a) but its not right.

2. Originally Posted by ardam
Express in the form a + bi, where a and b are real

(a) sin(3pi/2 - i)

(b) cosh(1 + i)

have tried using addition formula on (a) but its not right.
Note that for $z\in\mathbb{C}$, $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

Thus, $\sin\left(\tfrac{3\pi}{2}-i\right)=\frac{e^{\frac{3\pi}{2}i+1}-e^{-\frac{3\pi}{2}i-1}}{2i}$ $=\frac{e\left(\cos\left(\tfrac{3\pi}{2}\right)+i\s in\left(\tfrac{3\pi}{2}\right)\right)-e^{-1}\left(\cos\left(-\tfrac{3\pi}{2}\right)+i\sin\left(-\tfrac{3\pi}{2}\right)\right)}{2i}=\frac{-ei-e^{-1}i}{2i}$ $=-\frac{e+e^{-1}}{2}=\color{red}\boxed{-\cosh(1)}$

Does this make sense?

Try part (b). Note that if $z\in\mathbb{C}$, then $\cosh(z)=\frac{e^z+e^{-z}}{2}$.

3. Originally Posted by Chris L T521
Note that for $z\in\mathbb{C}$, $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

Thus, $\sin\left(\tfrac{3\pi}{2}-i\right)=\frac{e^{\frac{3\pi}{2}i+1}-e^{-\frac{3\pi}{2}i-1}}{2i}$ $=\frac{e\left(\cos\left(\tfrac{3\pi}{2}\right)+i\s in\left(\tfrac{3\pi}{2}\right)\right)-e^{-1}\left(\cos\left(-\tfrac{3\pi}{2}\right)+i\sin\left(-\tfrac{3\pi}{2}\right)\right)}{2i}=\frac{-ei-e^{-1}i}{2i}$ $=-\frac{e+e^{-1}}{2}=\color{red}\boxed{-\cosh(1)}$

Does this make sense?

Try part (b). Note that if $z\in\mathbb{C}$, then $\cosh(z)=\frac{e^z+e^{-z}}{2}$.
cheers for that.
right so cosh(z)

well sub 1 + i in

$\cosh(z)=\frac{e^{1+i} +e^{-1-i}}{2}$

then can you take e outside bracket or treat e^1+i as e^i.e?

5. Originally Posted by ardam
$\sin (z) = \sin (x)\cosh (y) + i\cos (x)\sinh (y)$
$\cos (z) = \cos (x)\cosh (y) - i\sin (x)\sinh (y)$
$\sinh (z) = \sinh (x)\cos (y) + i\cosh (x)\sin (y)$
$\cosh (z) = \cosh (x)\cos (y) - i\sinh (x)\sin (y)$