# The Square and Square Root Functions of a Complex Variable ... Gamelin, Ch. 1, Section 4

#### Peter

##### Well-known member
MHB Site Helper
I am reading Theodore W. Gamelin's book: "Complex Analysis" ...

I am focused on Chapter 1: The Complex Plane and Elementary Functions ...

I am currently reading Chapter 1, Section 4: The Square and Square Root Functions ... and need some help in verifying a remark by Gamelin ... ...

The relevant section from Gamelin is as follows:

In the above text by Gamelin we read the following .... ...

" ... ... Every value $$\displaystyle w$$ in the slit plane is the image of exactly two $$\displaystyle z$$ values. one in the (open) right half-plane [Re $$\displaystyle z \gt 0$$], the other in the left half-plane [Re $$\displaystyle z \lt 0$$]. ... ... "

Now, I wanted to demonstrate via an example that a value of $$\displaystyle w$$ was given by two values of $$\displaystyle z$$ ... so I let $$\displaystyle w = 1 + i$$ ... and proceeded as follows ...

$$\displaystyle w = 1 + i$$

so that

$$\displaystyle w = 2^{ \frac{1}{2} } e^{ i \frac{ \pi }{ 4} }$$

So then we have ... ...

$$\displaystyle z_1 = f_1(w) = w^{ \frac{1}{2} } = 2^{ \frac{1}{4} } e^{ i \frac{ \pi }{8} }$$

... and ...

$$\displaystyle z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} }$$

(Note that Gamelin uses $$\displaystyle f_2(w)$$ for the second branch of $$\displaystyle w^{ \frac{1}{2} }$$ ... and, further, notes that $$\displaystyle f_2(w) = - f_1(w)$$ ... ... ... ... )

My problem is that I do not believe my value or $$\displaystyle z_2$$ is correct ... but I cannot see where my process for calculating $$\displaystyle z_2$$ is wrong ...

Can someone please explain my mistake and show and explain the correct process for calculating $$\displaystyle z_2$$ ... ...

Help will be appreciated ...

Peter

Last edited:

#### HallsofIvy

##### Well-known member
MHB Math Helper
What you have is correct. $$\omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}}$$ so we have $$z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}$$. We can also write $$\omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}}$$ so that $$z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}$$. Since $$\frac{9\pi}{8}= \pi+ \frac{\pi}{8}$$, $$e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}}$$ because $$e^{i\pi}= -1$$.

#### Peter

##### Well-known member
MHB Site Helper
What you have is correct. $$\omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}}$$ so we have $$z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}$$. We can also write $$\omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}}$$ so that $$z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}$$. Since $$\frac{9\pi}{8}= \pi+ \frac{\pi}{8}$$, $$e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}}$$ because $$e^{i\pi}= -1$$.

Thanks for the help, HallsofIvy ....

However I still think my calculation of $$\displaystyle z_2$$ ... that is $$\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} }$$ is incorrect ...

Note that $$\displaystyle e^{ - i \frac{ \pi }{8} } \neq - e^{ i \frac{ \pi }{8} }$$ ...

But ... your post showed me the way ... as follows ..

$$\displaystyle z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = -2^{ \frac{1}{4} } e^{ i \frac{ \pi }{8} }$$

So ...

$$\displaystyle z_2 = 2^{ \frac{1}{4} } ( - e^{ i \frac{ \pi }{8} } ) = 2^{ \frac{1}{4} } e^{ i \frac{ 9 \pi }{8} }$$

or ... if you want the argument to be between $$\displaystyle - \pi$$ and $$\displaystyle + \pi$$ ...

$$\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ 7 \pi }{8} }$$

Is that correct now ... ?

Peter

#### Opalg

##### MHB Oldtimer
Staff member
$$\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ 7 \pi }{8} }$$

Is that correct now ... ?

Peter
That is correct. Notice that $e^{ - i \frac{ 7 \pi }{8} } = e^{ -i \pi + i\frac{ \pi }{8}} = e^{-i\pi}e^{i\frac\pi8} = -e^{i\frac\pi8}$. Thus $z_2 = -z_1$, as must always be the case with square roots.