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a) Re(iz)=-Im(z) b) Im(iz)=Re(z) c) (z+1)^2=z^2+2z+1 3. Do the following operations and simplify your answer. (1+2i)/(3-4i)+(2-i)/(5i) (5i)/((1-i)(2-i)(3-i)) c) (1-i)^3

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WallaceAdvanced · Tutor for 1 years

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### The simplified answers are (a) $\frac{2}{5i}$, (b) $\frac{5i}{1}$, (c) 2 - 2i

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## Step 1: Verify equations a and b ### For equation $a $, $ \Re(i z) = -\Im(z) $, if $ z = x + yi $, then $ i z = i(x + yi) = -y + xi $; thus, $\Re(i z) = -y $ and $\Im(z) = y$. Therefore, $\Re(i z) = -\Im(z)$ holds. ### For equation $b $, $ \Im(i z) = \Re(z) $, if $ z = x + yi $, then $ i z = i(x + yi) = -y + xi $; thus, $\Im(i z) = x $ and $\Re(z) = x $. Therefore, $ \Im(i z) = \Re(z)$ holds. ## Step 2: Verify equation c ### Expand the left-hand side $(z+1)^2 = (z+1)(z+1) = z^2 + 2z +1 $. Both sides are equal, thus $(z+1)^2 = z^2 + 2z +1 $ holds. ## Step 3: Simplify part (a) ### Rationalize the denominators and combine fractions: \[ \frac{1+2i}{3-4i} \cdot \frac{3+4i}{3+4i} = \frac{(1+2i)(3+4i)}{3^2+(4i)^2} = \frac{3 + 4i + 6i - 8}{9 + 16} = \frac{-5 + 10i}{25} = -\frac{1}{5} + \frac{2i}{5} \] \[ \frac{2-i}{5i} = \frac{2-i}{5i} \cdot \frac{-i}{-i} = \frac{(-2i) + (-1)}{- 25} = \frac{-1}{-25} - \frac{2i}{-25} = \frac{1}{5} + \frac{2}{5i} \] ### Add the two results: \[ -\frac{1}{5} + \frac{2i}{5} + \frac{1}{5} + \frac{2}{5i} = \frac{2i}{5} \] ## Step 4: Simplify part (b) ### Factorize the denominator and cancel as much as possible: \[ \frac{5i}{(1-i)(2-i)(3-i)} = \frac{5i}{(1-i)} \cdot \frac{1}{(2-i)} \cdot \frac{1}{(3-i)} = \frac{5i}{1} \] ## Step 5: Simplify part (c) ### Expand and simplify: \[ (1-i)^3 = (1-i)(1-i)(1-i) = (1-i)^2 (1-i) = (1-2i+i^2) (1-i) = (1-2i-1) (1-i) = -2i(1-i) = -2i + 2 \]

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a) Re(iz)=-Im(z) b) Im(iz)=Re(z) c) (z+1)^2=z^2+2z+1 3. Do the following operations and simplify your answer. (1+2i)/(3-4i)+(2-i)/(5i) (5i)/((1-i)(2-i)(3-i)) c) (1-i)^3

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