Studdy Solution
STEP 1
What is this asking?
We need to show that the inverse of the conjugate of z is the same as the conjugate of the inverse of z, assuming z isn't zero.
Watch out!
Remember that the conjugate flips the sign of the imaginary part, and the inverse flips the complex number "upside down" in a special way.
Don't mix them up!
STEP 2
1. Define the complex number and its conjugate.
2. Compute the left-hand side.
3. Compute the right-hand side.
4. Compare the two sides.
STEP 3
Let's **define** our complex number z as z=a+bi, where a and b are real numbers, and i is the imaginary unit.
STEP 4
The **conjugate** of z, denoted as zˉ, is found by changing the sign of the imaginary part.
So, zˉ=a−bi.
STEP 5
We want to compute zˉ−1.
First, let's find the inverse of zˉ.
We multiply zˉ by zˉˉzˉˉ, which is just a fancy way of multiplying by one.
STEP 6
zˉ−1=zˉ1=a−bi1⋅a+bia+bi=a2−(bi)2a+bi
Remember, i2=−1, so (bi)2=−b2.
STEP 7
zˉ−1=a2−(−b2)a+bi=a2+b2a+bi=a2+b2a+a2+b2bi
So, we have our **left-hand side**!
STEP 8
Now, let's compute (z−1).
First, we find z−1.
STEP 9
z−1=z1=a+bi1⋅a−bia−bi=a2+b2a−bi=a2+b2a−a2+b2bi
STEP 10
Now, we take the **conjugate** of this result:
(z−1)=a2+b2a−a2+b2bi=a2+b2a+a2+b2bi
And there's our **right-hand side**!
STEP 11
Let's compare the two sides:
zˉ−1=a2+b2a+a2+b2bi
(z−1)=a2+b2a+a2+b2biThey're the same!
STEP 12
We've shown that zˉ−1=(z−1)!
We did this by carefully calculating both sides and seeing they match.
