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Which of the following is equal to $e^{ar{z}}$, where $z = x + iy$ with $x$ and $y$ real numbers? A - HSC - SSCE Mathematics Extension 2 - Question 8 - 2024 - Paper 1

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Which-of-the-following-is-equal-to-$e^{ar{z}}$,-where-$z-=-x-+-iy$-with-$x$-and-$y$-real-numbers?-A-HSC-SSCE Mathematics Extension 2-Question 8-2024-Paper 1.png

Which of the following is equal to $e^{ar{z}}$, where $z = x + iy$ with $x$ and $y$ real numbers? A. $\bar{e}$ B. $e^{-z}$ C. $e^{2x}e^{y}$ D. $e^{-2x}e^{\bar{z}}$

Worked Solution & Example Answer:Which of the following is equal to $e^{ar{z}}$, where $z = x + iy$ with $x$ and $y$ real numbers? A - HSC - SSCE Mathematics Extension 2 - Question 8 - 2024 - Paper 1

Step 1

Identify the expression: $e^{ar{z}}$

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Answer

Given that z=x+iyz = x + iy, the conjugate of zz, denoted as ar{z}, is xiyx - iy. Hence, we need to find e^{ar{z}} = e^{x - iy}.

Step 2

Use the properties of exponents

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Answer

Using the exponent rule, we can split this into: e^{ar{z}} = e^x imes e^{-iy}.

Step 3

Apply Euler's formula for $e^{-iy}$

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Answer

According to Euler's formula, we have: e^{-iy} = rac{1}{e^{iy}} = rac{1}{ ext{cos}(y) + i ext{sin}(y)} = ext{cos}(y) - i ext{sin}(y). Therefore, we combine this with exe^x to express: e^{ar{z}} = e^x( ext{cos}(y) - i ext{sin}(y)).

Step 4

Compare with the given options

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Answer

Now we analyze the answer choices:

  • A: eˉ\bar{e} does not apply here.
  • B: ez=exiy=ex(eiy)e^{-z} = e^{-x - iy} = e^{-x}(e^{-iy}), which is incorrect.
  • C: e2xeye^{2x}e^{y} does not match either.
  • D: e2xezˉe^{-2x}e^{\bar{z}} does not correspond. Thus, the correct choice is A, eˉ\bar{e}.

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