Solve the complex number problems.

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Understand the Problem

The image contains multiple choice questions related to complex numbers, which require applying various properties and operations to solve them. Here's a breakdown of each question:

  1. Calculate the value of a complex number expression raised to a power.
  2. Find the value of an expression involving squares, given that a complex number equals the square root of a fraction with complex numbers.
  3. Determine the possible values of x, given that two complex numbers are conjugates.
  4. Determine the value of z, given an equation involving the square of z and the square of iz.
  5. Compute the value of i raised to the power of -57.
  6. Find the complex number, given its conjugate.

We will classify these as math questions.

Answer

1. $2i$ 2. $\frac{a^2+b^2}{c^2+d^2}$ 3. no value of x. 4. z = 0 5. -i 6. $\frac{-1}{i+1}$
Answer for screen readers
  1. (b) $2i$
  2. (a) $\frac{a^2+b^2}{c^2+d^2}$
  3. (d) no value of x.
  4. (d) z = 0
  5. (a) -i
  6. (b) $\frac{-1}{i+1}$

Steps to Solve

  1. Calculate $(\frac{2i}{1+i})^2$

Simplify the expression inside the parenthesis first by multiplying the numerator and denominator by the conjugate of the denominator: $$ \frac{2i}{1+i} = \frac{2i(1-i)}{(1+i)(1-i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i + 2}{1 + 1} = \frac{2 + 2i}{2} = 1 + i $$ Now, square the result: $$ (1+i)^2 = (1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i $$

  1. Find $(x^2 + y^2)^2$ given $x + iy = \sqrt{\frac{a+ib}{c+id}}$

First, find the magnitude squared of $x+iy$: $$ |x+iy|^2 = x^2 + y^2 $$ Given that $x + iy = \sqrt{\frac{a+ib}{c+id}}$, then $$ (x+iy)^2 = \frac{a+ib}{c+id} $$ Then, $$ |(x+iy)^2| = |x+iy|^2 = |\frac{a+ib}{c+id}| = \frac{|a+ib|}{|c+id|} = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} $$ So, $$ x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} $$ Squaring both sides: $$ (x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2} $$

  1. Find x when $\sin x + i\cos 2x$ and $\cos x - i\sin 2x$ are conjugates

For two complex numbers to be conjugates, their real parts must be equal, and their imaginary parts must be the negative of each other. Therefore, $$ \sin x = \cos x $$ $$ \cos 2x = \sin 2x $$ From $\sin x = \cos x$, we have $x = (n + \frac{1}{4})2\pi = 2n\pi + \frac{\pi}{4}$ or $x = n\pi + \frac{\pi}{4} $, where $n$ is an integer. From $\cos 2x = \sin 2x$, $2x = n\pi + \frac{\pi}{4}$ or $x = n\frac{\pi}{2} + \frac{\pi}{8}$, where $n$ is an integer.

If $x = n\frac{\pi}{2} + \frac{\pi}{8}$, $\sin(n\frac{\pi}{2} + \frac{\pi}{8}) = \cos(n\frac{\pi}{2} + \frac{\pi}{8})$. This implies that $x = \frac{\pi}{4}$. If we substitute $x = \frac{\pi}{4}$, $\sin x = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ and $\cos x = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Also, $\cos 2x = \cos \frac{\pi}{2} = 0 $ and $\sin 2x = \sin \frac{\pi}{2} = 1 $. Since $\cos 2x = \sin 2x$ is not possible, then there is no solution.

  1. If $z^2 = (iz)^2$, then find z

Given $z^2 = (iz)^2$, we have $z^2 = i^2 z^2 = -z^2$, so $z^2 = -z^2$ $2z^2 = 0 $, therefore $z^2 = 0$ and $z=0$.

  1. Calculate $i^{-57}$

$i^{-57} = \frac{1}{i^{57}} = \frac{1}{i^{56} * i} = \frac{1}{(i^4)^{14} * i} = \frac{1}{1^{14} * i} = \frac{1}{i} = \frac{1}{i} * \frac{-i}{-i} = \frac{-i}{-i^2} = \frac{-i}{1} = -i$

  1. Find the complex number whose conjugate is $\frac{1}{i-1}$

If the conjugate of a complex number $z$ is $\frac{1}{i-1}$, then $z = \overline{(\frac{1}{i-1})} = \frac{1}{\overline{(i-1)}} = \frac{1}{-i-1} = \frac{1}{-(i+1)} = \frac{-1}{i+1}$.

  1. (b) $2i$
  2. (a) $\frac{a^2+b^2}{c^2+d^2}$
  3. (d) no value of x.
  4. (d) z = 0
  5. (a) -i
  6. (b) $\frac{-1}{i+1}$

More Information

Complex numbers are represented in the form a + bi, where a and b are real numbers, and i is the imaginary unit, defined as the square root of -1.

Tips

  1. When squaring a complex number like $(1+i)^2$ forgetting the cross term $2i$.
  2. Not rationalizing the denominator when dividing complex numbers.
  3. Confusing a complex number with its conjugate. The conjugate of $a + bi$ is $a - bi$.
  4. Not knowing the powers of $i$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.
  5. When x+iy = sqrt((a+ib)/(c+id)), forgetting to take square of both sides before finding the magnitude.

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