# Is there any better way of finding the required value

If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$

The solution that I have is this $$z=\cos\theta+i\sin\theta \implies$$ $$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}$$ $$=\frac{2\cos^2\frac{\theta}{2}+i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}-i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\cdot \frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$ $$=\cot\frac{\theta}{2}\cdot\frac{\color{red}{i}}{\color{red}{i}}×\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$ $$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\color{red}{i}\sin\frac{\theta}{2}-\color{red}{i}i\cos\frac{\theta}{2}}$$ $$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\color{red}{i}\sin\frac{\theta}{2}}$$ $$=\color{red}{i}\cot\frac{\theta}{2}$$ Now how on earth one will imagine the steps written in red. I am looking for an easy and a logical answer to this question. The solution that I have is impractical as you all can see. I tried it doing by $$e^{i\theta}$$ but no good happen.

Any help is greatly appreciated.

• That step seems highly natural to me. If you look at the fraction, it is obvious that the denominator is just the numerator with the real and imaginary parts reversed, and the imaginary part negated - exactly the effect of dividing a complex number by $i$. To undo that, multiply the denominator by $i$. And if you multiply the denominator by something, you have to also multiply the numerator by the same thing. This isn't pulled out of thin air. It is an obvious step to take if you have experience with multiplying by $i$ and $-i$. Nov 23, 2022 at 14:17
• This is not disagreeing that David Quinn's method - which is just the standard technique for dividing complex numbers - is better, though. Nov 23, 2022 at 14:25

Rather than have to guess to use a special trick like multiplying by $$\frac{i}{i}$$, just multiply top and bottom by the conjugate of the denominator. This can be done in trig form, or exponential form, or just like this: $$\frac{1+z}{1-z}\times\frac{1-\bar{z}}{1-\bar{z}}$$ $$=\frac{1+z-\bar{z}-z\bar{z}}{1-z-\bar{z}+z\bar{z}}$$ $$=\frac{1+2i\sin\theta-1}{1-2\cos\theta+1}$$

Now use the half-angles: $$=\frac{4i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{4\sin^2\frac{\theta}{2}}$$ $$=i\cot\frac{\theta}{2}$$

• Still you actually multiplied by 1 Nov 19, 2022 at 16:41
• Yes, but multiplying top and bottom of a fraction of complex numbers by the conjugate of the denominator is a standard technique, known as realizing the denominator, which is used in the basic algebra of complex numbers and introduced at the elementary stage. I wouldn’t call it a special trick in that sense. Nov 19, 2022 at 17:12
• Realizing the denominator or rationalizing the denominator $?$ Nov 19, 2022 at 17:56
• Realizing, as in “making it real”. This is analogous to rationalizing the denominator in the case of fractions of surds Nov 19, 2022 at 18:53

Observe that $$\dfrac{\cos t+i\sin t}{\cos t-i\sin t}=\cos2t+i\sin2t$$

If $$\dfrac z1=\cos2t+i\sin2t=\dfrac{\cos t+i\sin t}{\cos t-i\sin t}$$

$$\dfrac{1+z}{1-z}=\cdots=?$$

• Can you please elaborate$?$ Nov 19, 2022 at 16:28
• @Vanessa, Please share the point of confusion Dec 1, 2022 at 11:25

Observe that $$\begin{cases}\cos(t)=\dfrac {e^{it}+e^{-it}}2\\ \sin(t)=\dfrac{e^{it}-e^{-it}}{2i}\end{cases}\quad$$ leads to $$\tan(t)=\dfrac 1i\cdot\dfrac{e^{it}-e^{-it}}{e^{it}+e^{-it}}$$

Here you are given $$z=e^{i\theta}$$, therefore the expression $$\dfrac{1+z}{1-z}=\dfrac{1+e^{i\theta}}{1-e^{i\theta}}$$

Considering the formula for $$\tan(t)$$ given above, you want in fact to make the half angle appear by factoring $$e^{i\theta/2}$$.

$$\dfrac{e^{i\theta/2}\Big(e^{-i\theta/2}+e^{i\theta/2}\Big)}{e^{i\theta/2}\Big(e^{-i\theta/2}-e^{i\theta/2}\Big)}$$

I agree this is still some magical appearance of a quantity in the style $$1=\frac xx$$, but I think it is a bit more natural than the solve you presented.

See it as introducing the middle point between $$0$$ and $$\theta$$, i.e. between $$1=e^{i0}$$ and $$e^{i\theta}$$ in order to "homogenize" the expression.

I don't think the question is really accurate in the first place, the value of $$\frac{1+z}{1-z}$$ can just be determined by plugging in the expression for $$z$$. To bring it in the final form is a matter of taste, I believe it mostly comes from the experience and knowing which result you want to get. Multiplying by $$1$$ and adding $$0$$ are among the most common tricks, but I don't think there exists any universal method.