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For z₁ = 5 - 2i and z₂ = -3 + 6i, find z₁ + z₂ and 2z₁ - z₂ in the form a + ib.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

For $z_1 = 5 - 2i$ and $z_2 = -3 + 6i$ , find $z_1 + z_2$ and $2z_1 - z_2$ in the form $a + ib$ .

Need a reminder? The key ideaShow

Add and take away part by part. Multiply out the brackets and put $-1$ for $i^2$ : $(a + ib)(c + id) = (ac - bd) + i(ad + bc)$ . The commutative, associative and distributive laws hold, and so do all the algebraic identities.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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← Work out √(-16) × √(-25) correctly. Then work out √((-16) × (-25)).Find (2 + 3i)(4 - i) and (1 - i)³ in the form a + ib. →

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For z₁ = 5 - 2i and z₂ = -3 + 6i, find z₁ + z₂ and 2z₁ - z₂ in the form a + ib.

The worked solution

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