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Given Z1 = 3 + 2i and Z2 = 2 − i, find Z1 + Z2, Z2 - Z1, 2Z1 + 3Z2, and Z1 Z2.

Mathematics

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Answers

6 parts
a

5+i\text{}5 + i\text{}

b

−1−3i\text{} -1 - 3i\text{}

c

12+i\text{}12 + i\text{}

d

8+i\text{}8 + i\text{}

e

45+75i\text{}\frac{4}{5} + \frac{7}{5}i\text{}

f

13\text{}\sqrt{13}\text{}

Step-by-step solution

17 steps

Given the complex numbers:
Z1=3+2iZ_1 = 3 + 2i
Z2=2−iZ_2 = 2 - i

We will find the required expressions.

Part a· 2 steps

Find Z1+Z2Z_1 + Z_2.

  1. Substitute the given values of Z1Z_1 and Z2Z_2.

    Z1+Z2=(3+2i)+(2−i)Z_1 + Z_2 = (3 + 2i) + (2 - i)

  2. Combine the real parts and the imaginary parts.

    Z1+Z2=(3+2)+(2i−i)Z_1 + Z_2 = (3 + 2) + (2i - i) Z1+Z2=5+iZ_1 + Z_2 = 5 + i

    The sum is 5+i\boxed{\text{}5 + i\text{}}

Part b

−1−3i\text{} -1 - 3i\text{}

Find Z2−Z1Z_2 - Z_1.

  1. Substitute the given values of Z1Z_1 and Z2Z_2.

    Z2−Z1=(2−i)−(3+2i)Z_2 - Z_1 = (2 - i) - (3 + 2i)

  2. Distribute the negative sign and combine the real parts and the imaginary parts.

    Z2−Z1=2−i−3−2iZ_2 - Z_1 = 2 - i - 3 - 2i Z2−Z1=(2−3)+(−i−2i)Z_2 - Z_1 = (2 - 3) + (-i - 2i) Z2−Z1=−1−3iZ_2 - Z_1 = -1 - 3i

    The difference is −1−3i\boxed{\text{} -1 - 3i\text{}}

Part c

12+i\text{}12 + i\text{}

Find 2Z1+3Z22Z_1 + 3Z_2.

  1. Multiply Z1Z_1 by 2 and Z2Z_2 by 3.

    2Z1=2(3+2i)=6+4i2Z_1 = 2(3 + 2i) = 6 + 4i 3Z2=3(2−i)=6−3i3Z_2 = 3(2 - i) = 6 - 3i

  2. Add the results.

    2Z1+3Z2=(6+4i)+(6−3i)2Z_1 + 3Z_2 = (6 + 4i) + (6 - 3i)

  3. Combine the real parts and the imaginary parts.

    2Z1+3Z2=(6+6)+(4i−3i)2Z_1 + 3Z_2 = (6 + 6) + (4i - 3i) 2Z1+3Z2=12+i2Z_1 + 3Z_2 = 12 + i

    The result is 12+i\boxed{\text{}12 + i\text{}}

Part d

8+i\text{}8 + i\text{}

Find Z1Z2Z_1 Z_2.

  1. Substitute the given values of Z1Z_1 and Z2Z_2.

    Z1Z2=(3+2i)(2−i)Z_1 Z_2 = (3 + 2i)(2 - i)

  2. Multiply the complex numbers using the distributive property (FOIL).

    Z1Z2=3(2)+3(−i)+2i(2)+2i(−i)Z_1 Z_2 = 3(2) + 3(-i) + 2i(2) + 2i(-i) Z1Z2=6−3i+4i−2i2Z_1 Z_2 = 6 - 3i + 4i - 2i^2

  3. Substitute i2=−1i^2 = -1 and simplify.

    Z1Z2=6+i−2(−1)Z_1 Z_2 = 6 + i - 2(-1) Z1Z2=6+i+2Z_1 Z_2 = 6 + i + 2 Z1Z2=8+iZ_1 Z_2 = 8 + i

    The product is 8+i\boxed{\text{}8 + i\text{}}

Part e

45+75i\text{}\frac{4}{5} + \frac{7}{5}i\text{}

Find Z1Z2\frac{Z_1}{Z_2}.

  1. Substitute the given values of Z1Z_1 and Z2Z_2.

    Z1Z2=3+2i2−i\frac{Z_1}{Z_2} = \frac{3 + 2i}{2 - i}

  2. Multiply the numerator and denominator by the conjugate of the denominator.

    The conjugate of 2−i2 - i is 2+i2 + i.

    Z1Z2=(3+2i)(2+i)(2−i)(2+i)\frac{Z_1}{Z_2} = \frac{(3 + 2i)(2 + i)}{(2 - i)(2 + i)}

  3. Multiply the terms in the numerator.

    (3+2i)(2+i)=3(2)+3(i)+2i(2)+2i(i)(3 + 2i)(2 + i) = 3(2) + 3(i) + 2i(2) + 2i(i) =6+3i+4i+2i2= 6 + 3i + 4i + 2i^2 =6+7i+2(−1)= 6 + 7i + 2(-1) =6+7i−2= 6 + 7i - 2 =4+7i= 4 + 7i

  4. Multiply the terms in the denominator.

    (2−i)(2+i)=22−i2(2 - i)(2 + i) = 2^2 - i^2 =4−(−1)= 4 - (-1) =4+1= 4 + 1 =5= 5

  5. Combine the numerator and denominator and express in the form a+bia + bi.

    Z1Z2=4+7i5\frac{Z_1}{Z_2} = \frac{4 + 7i}{5} Z1Z2=45+75i\frac{Z_1}{Z_2} = \frac{4}{5} + \frac{7}{5}i

    The quotient is 45+75i\boxed{\text{}\frac{4}{5} + \frac{7}{5}i\text{}}

Part f

13\text{}\sqrt{13}\text{}

Find ∣Z1∣|Z_1|.

  1. Recall the formula for the modulus of a complex number z=a+biz = a + bi, which is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}.

    For Z1=3+2iZ_1 = 3 + 2i, we have a=3a = 3 and b=2b = 2.

  2. Substitute the values into the formula.

    ∣Z1∣=32+22|Z_1| = \sqrt{3^2 + 2^2} ∣Z1∣=9+4|Z_1| = \sqrt{9 + 4} ∣Z1∣=13|Z_1| = \sqrt{13}

    The modulus is 13\boxed{\text{}\sqrt{13}\text{}}

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