So I recently found out that

\[\sin {(a + bi)} = i\sinh {(b - ai)}\]

\[\cos {(a + bi)} = \cosh {(b - ai)}\]

and that

\[\sinh {(x)} = \frac {e^x - e^{-x}}{2}\]

\[\cosh {(x)} = \frac {e^x + e^{-x}}{2}\]

Since

\[\frac {a}{b} + \frac {b}{a} = \frac {a^2 + b^2}{ab}\]

\[\frac {a}{b} - \frac {b}{a} = \frac {a^2 - b^2}{ab}\]

and

\[(\frac {a}{b})^{-x} = (\frac {b}{a})^x\]

Re-arranging the equations for \(\sinh{(x)}\) and \(\cosh {(x)}\) gives us

\[\sinh {(x)} = \frac {e^{2x} - 1}{2e^x}\]

\[\cosh {(x)} = \frac {e^{2x} + 1}{2e^x}\]

Substituting these into \(\sin {(a + bi)}\) and \(\cos {(a + bi)}\) when \(x = b - ai\) gives us

\[\sin {(a + bi)} = i\frac {e^{2(b - ai)} - 1}{2e^{b - ai}}\]

\[\cos {(a + bi)} = \frac {e^{2(b - ai)} + 1}{2e^{b - ai}}\]

I'm going to try and expand and substitute by using

\[e^{a + bi} = e^a (\cos {(b)} + i\sin {(b)})\]

Let \(\cos {\theta} + i\sin {\theta} = \) \(cis\) \(\theta\). So

\[e^{a + bi} = e^a \cdot cis (b)\]

Next we substitute this new equation in to get

\[\sin {(a + bi)} = i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)}\]

\[\cos {(a + bi)} = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)}\]

So what does \(cis(a + bi)\) equal

Substituting in the two equations gives us

\[cis(a + bi) = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)} + i(i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)})\]

That looks unfriendly, let's simplify it

\[cis(a + bi) = \frac {1}{e^{b} \cdot cis(-a)}\]

(Go through the steps, it works)

This can be simplified even further

\[cis(a + bi) = \frac {cis(a)}{e^b}\]

P.s. \[(\cos {\theta} + i\sin {\theta})^n = \cos {\theta n} + i\sin{\theta n}\]

## Comments

Sort by:

TopNewest\[\begin{align*} \text{cis}(a+bi) &= e^{i(a+bi)}\\ &= e^{ia-b}\\ &= \dfrac{e^{ia}}{e^b}\\ &= \dfrac{\text{cis}(a)}{e^b} \end{align*}\] – Daniel Liu · 2 years, 9 months ago

Log in to reply

– Jack Rawlin · 2 years, 9 months ago

I did consider that but I wanted to make it as complex as possible so it would be fun to work out.Log in to reply

– B.S.Bharath Sai Guhan · 2 years, 9 months ago

Lol..Nice note!!Log in to reply

Very nice note. Learned a lot! – Trevor Arashiro · 2 years, 9 months ago

Log in to reply

Although trigonometric functions can operate complex numbers, I think there is no complex angle but only real angles. Just like a doctor who can save lives, not all lives being saved are human being. Do you agree with me? – Lu Chee Ket · 2 years, 7 months ago

Log in to reply