Linear combination of complex

If z1=5+3i and z2=4-2i, write the following in the form a+bi

a) 4z1+6z2
b) z1*z2

Result

x =  44
y = (Correct answer is: 26+2i)

Solution:

$$\begin{gathered} z_1=5+3i \\ {z}_2=4-2i \\ x=4\cdot z_1+6\cdot z_2 \\ x=4\cdot (5+3i)+6\cdot (4-2i) \\ x=20+12i+24-12i \\ x=(20+24)+(12-12)i \\ x=44 \end{gathered}$$

$$\begin{gathered} y=z_1\cdot z_2 \\ y=(5+3i)\cdot (4-2i) \\ 5\cdot 4+5\cdot (-2i)+3i\cdot 4+3i\cdot (-2i) \\ y=20-10i+12i-6i^2 \\ y=20-10i+12i+6 \\ y=20+6+i(-10+12) \\ y=26+2i \end{gathered}$$

Try another example

Showing 1 comment:

Alena

for x see calculation:

https://www.hackmath.net/en/calculator/complex-number?input=4%285%2B3i%29%2B6%284-2i%29

for y:

https://www.hackmath.net/en/calculator/complex-number?input=%285%2B3i%29%2A%284-2i%29

2 months ago  Like

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