Jack Ingram

2022-10-21

I am doing some matrix multiplication, and at one point it is stated that,

$$\left(\begin{array}{cc}1& 0\\ 0& {e}^{\frac{i\pi}{4}}\end{array}\right)\left(\begin{array}{c}1\\ i\end{array}\right)\text{}=\left(\begin{array}{c}1\\ {e}^{i\frac{3\pi}{4}}\end{array}\right)$$

but how does $i{e}^{i\frac{\pi}{4}}={e}^{i\frac{3\pi}{4}}$?

I have tried taking Euler's formula but I can only get it to reduce down

$$\frac{-i-1}{\sqrt{2}i}$$

I have also tried converting to polar form, but I just really can't see how these two are equivalent.

$$\left(\begin{array}{cc}1& 0\\ 0& {e}^{\frac{i\pi}{4}}\end{array}\right)\left(\begin{array}{c}1\\ i\end{array}\right)\text{}=\left(\begin{array}{c}1\\ {e}^{i\frac{3\pi}{4}}\end{array}\right)$$

but how does $i{e}^{i\frac{\pi}{4}}={e}^{i\frac{3\pi}{4}}$?

I have tried taking Euler's formula but I can only get it to reduce down

$$\frac{-i-1}{\sqrt{2}i}$$

I have also tried converting to polar form, but I just really can't see how these two are equivalent.

Martha Dickson

Beginner2022-10-22Added 20 answers

This is the whole process of solving:

$$i{e}^{i\frac{\pi}{4}}={e}^{i\frac{\pi}{2}}{e}^{i\frac{\pi}{4}}={e}^{i\frac{3\pi}{4}}$$

$$i{e}^{i\frac{\pi}{4}}={e}^{i\frac{\pi}{2}}{e}^{i\frac{\pi}{4}}={e}^{i\frac{3\pi}{4}}$$

Kamila Frye

Beginner2022-10-23Added 4 answers

As ${i}^{4}=1$, i represents a rotation by $2\pi /4=\pi /2$ radians counterclockwise . Similarly, ${e}^{i\pi /4}$ represents a rotation by another $\pi /4$ radians multiplying represents the composition of the two transformations, for a total of $3\pi /4$ radians or ${e}^{3i\pi /4}$

Remember that ${e}^{it}$ is a rotation of t radians due to Euler's identity ${e}^{it}=\mathrm{cos}t+i\mathrm{sin}t$

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