Problems from Operations with complex numbers in binomic form

Let's calculate:

  1. $$(3+2i)+(8-2i)$$
  2. $$(3+2i)-(8-2i)$$
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Development:

  1. $$(3+2i)+(8-2i)=(3+8)+(2+(-2))i=11+0i=11$$

    We have added the real parts $$(3 +8)$$ and we have added the imaginary parts $$(2 + (-2))$$. This last component is zero and therefore the solution is a real number.

  2. $$(3+2i)-(8-2i)=(3-8)+(2-(-2))i=-5+(2+2)i=-5+4i$$

    We substract real parts $$(3-8)$$ and the imaginary parts $$(2 - (-2))$$. The real part result is $$-5$$ and the imaginary part result is $$4$$.

Solution:

  1. $$11$$
  2. $$-5+4i$$
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Solve the following operations:

  1. $$(3-6i)\cdot(1+2i)=$$
  2. $$(2+i)\cdot(2+3i)\cdot(4-6i)=$$
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Development:

  1. With the formula of the product of complex numbers in binomic form: $$$ \displaystyle \begin{array}{rl} (3-6i)\cdot(1+2i) &= (3\cdot 1+6\cdot2) +(3\cdot2-6\cdot1)i= (3+12)+(6-6)i \\ & =15+0i= 15 \end{array} $$$
  2. We will first do the product of the first two complex numbers and then multiply the result by the third complex number.

    $$(2+i)\cdot(2+3i)= (2\cdot2-3\cdot1)+(2\cdot3+1\cdot2)i=1+8i$$

    $$ \displaystyle \begin{array}{rl} (1+8i)\cdot(4-6i)=&(1\cdot4-8\cdot(-6))+(8\cdot4+1\cdot(-6))\cdot and =\\ =&4+48+(32-6)\cdot i=52+26i \end{array}$$

Solution:

  1. $$15$$
  2. $$52+26i$$
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  1. Give the conjugate and the opposite of the following complex numbers: $$1-4i$$, $$-9-5i$$.
  2. Calculate $$(-11+29i):(2+3i)=$$
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Development:

  1. $$$ \displaystyle z=1-4i \Rightarrow \left\{ \begin{array}{l} \bar{z}=1-(-4)i=1+4i \\ -z=-(1-4i)=-1+4i \end{array} \right. $$$

    The first one corresponds to the conjugate and the second one to the opposite complex number. $$$ \displaystyle z=-9-5i \Rightarrow \left\{ \begin{array}{l} \bar{z}=-9-(-5)i=-9+5i \\ -z=-(-9-5i)=9+5i \end{array} \right. $$$

    The same as in the previous lines.

  2. Now, multiplying by the conjugate of $$2 + 3i$$ (which is $$2-3i$$) we have: $$$\dfrac{-11+29i}{2+3i}=\dfrac{-11+29i}{2+3i}\cdot \dfrac{2-3i}{2-3i}$$$ We do both products: $$$\dfrac{-11+29i}{2+3i}\cdot \dfrac{2-3i}{2-3i}=\dfrac{(-11\cdot2-29\cdot3)+(-11\cdot3+29\cdot2)i}{2^2+3^2}$$$ Joining terms and adding: $$$ \dfrac{(-11\cdot2-29\cdot3)+(-11\cdot3+29\cdot2)i}{2^2+3^2}=\dfrac{65+91i}{4+9}$$$ If we separate the fraction in two terms we obtain: $$$\dfrac{-11+29i}{2+3i}=\dfrac{65}{13}+\dfrac{91}{13}i$$$ Simplified, that will be: $$$\dfrac{-11+29i}{2+3i}=\dfrac{65}{13}+\dfrac{91}{13}i=5+7i$$$

Solution:

  1. For the first one:

    $$\bar{z}=1-(-4)i=1+4i$$

    $$-z=-(1-4i)=-1+4i$$

    For the second one:

    $$\bar{z}=-9-(-5)i=-9+5i$$

    $$-z=-(-9-5i)=9+5i$$

  2. $$5+7i$$
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