In the following questions, what is the value of z_1.z_2 ?
\left(3-i\right)\times\left(2+5i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=3-i
- z_2=2+5i
\left(3-i\right)\times\left(2+5i\right)=6-2i+15i-5i^2
Since i^2=-1, we have:
\left(3-i\right)\times\left(2+5i\right)=6+5+13i
\left(3-i\right)\times\left(2+5i\right)=11+13i
\left(1-2i\right) \times \left(1 + 2i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=1-2i
- z_2=1+2i
\left(1-2i\right) \times \left(1 + 2i\right)=1-2i+2i-4i^2
Since i^2=-1, we have:
\left(1-2i\right) \times \left(1 + 2i\right)=1+4=5
\left(1-2i\right) \times \left(1 + 2i\right)=5
\left(1-4i\right) \times \left(2 + i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=1-4i
- z_2=2+i
\left(1-4i\right) \times \left(2 + i\right)=2+i-8i-4i^2
Since i^2=-1, we have:
\left(1-4i\right) \times \left(2 + i\right)=2+4+i-8i=6-7i
\left(1-4i\right) \times \left(2 + i\right)=6-7i
\left(2-3i\right) \times \left(5 - 6i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=2-3i
- z_2=5-6i
\left(2-3i\right) \times \left(5 - 6i\right)=10-12i-15i+18i^2
Since i^2=-1, we have:
\left(2-3i\right) \times \left(5 - 6i\right)=10-18-27i=-8-27i
\left(2-3i\right) \times \left(5 - 6i\right)=-8-27i
\left(5+7i\right) \times \left(6 - 2i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=5+7i
- z_2=6-2i
\left(5+7i\right) \times \left(6 - 2i\right)=30-10i+42i-14i^2
Since i^2=-1, we have:
\left(5+7i\right) \times \left(6 - 2i\right)=30+14-10i+42i=44+32i
\left(5+7i\right) \times \left(6 - 2i\right)=44+32i
\left(3-9i\right) \times \left(2 - 8i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=3-9i
- z_2=2-8i
\left(3-9i\right) \times \left(2 - 8i\right)=6-24i-18i+72i^2
Since i^2=-1, we have:
\left(3-9i\right) \times \left(2 - 8i\right)=6-24i-18i-72=-66-42i
\left(3-9i\right) \times \left(2 - 8i\right)=-66-42i
\left(3-7i\right) \times \left(3 +7i\right)
Let z_1=a_1+b_1i and z_2=a_2+b_2i
Then:
z_1\times z_2=\left(a_1+b_1i\right)\times\left(a_2+b_2i\right)
Here, we have:
- z_1=3-7i
- z_2=3+7i
\left(3-7i\right) \times \left(3 +7i\right)=9+21i-21i-49i^2
Since i^2=-1, we have:
\left(3-7i\right) \times \left(3 +7i\right)=9+49=58
\left(3-7i\right) \times \left(3 +7i\right)=58