How do you multiply $(2 + i) (2 - 3 i)$ $(2+i)(2-3i)$ in trigonometric form?

Question

How do you multiply $(2 + i) (2 - 3 i)$ $(2+i)(2-3i)$ in trigonometric form?

1 Answer

Impact of this question

1797 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-21T17:13:59

$7 - 4 i$ $7-4i$

Explanation:

First, we have to know that
$i^{1} = i$ $i^1=i$
$i^{2} = - 1$ $i^2=-1$
$i^{3} = - i$ $i^3=-i$
$i^{4} = 1$ $i^4=1$

Then, we can expend the expression and simplify it with the above rule:

$(2 + i) (2 - 3 i)$ $(2+i)(2-3i)$
$= 4 - 6 i + 2 i - 3 i^{2}$ $=4-6i+2i-3i^2$
$= 4 - 4 i - 3 i^{2}$ $=4-4i-3i^2$
$= 4 - 4 i - 3 (- 1)$ $=4-4i-3(-1)$
$= 4 - 4 i + 3$ $=4-4i+3$
$= 7 - 4 i$ $=7-4i$

It's the answer! Hope it can help you :)

Science

Math

Humanities

... and beyond

How do you multiply $(2 + i) (2 - 3 i)$ $(2+i)(2-3i)$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (2+i)(2-3i) (2+i)(2−3i) (2+i)(2-3i) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply $(2 + i) (2 - 3 i)$ $(2+i)(2-3i)$ in trigonometric form?