Showing posts with label Complex numbers. Show all posts
Showing posts with label Complex numbers. Show all posts

Sunday, July 9, 2023

Factorise $p(z)=z^5+z^4-8z^3-28z^2+16z+96$ into linear factors.

 The original question was that given $-2-2i$ is a root and that $3$ other roots were real, factorise into linear factors. Since $p(z)$ has real coefficients this means the conjugate root is also a root, so both $-2-2i$ and $-2+2i$ are roots. 

Here we assume that there are $3$ real roots and go from there.



Saturday, July 8, 2023

Thursday, July 6, 2023

Monday, July 3, 2023

Express $\cos^5t$ in terms of $\cos 5t,\cos3t,\cos t$ and express $\sin^4t$ in terms of $\cos 4t, \cos 2t$

  • Other examples like this one are contained in the Algebra Notes Ch3 page 98 Example 3 and 4 ! Please also read this book, as it has many useful examples!!
  • RTB - Read the Book 😎











Thursday, June 15, 2023

Challenge:Complex reducible to quadratic equation: Solve $$z^4+2\left(a^2-b^2\right)z^2+\left(a^2+b^2\right)^2=0$$ where $a,b\in\mathbb{R}$

Solve $$z^4+2\left(a^2-b^2\right)z^2+\left(a^2+b^2\right)^2=0$$ where $a,b$ are real numbers.

Hint 1 : Put $w=z^2$ and find $w$ first.  Then for each $w$, solve $z^2=w$.

Hint 2: see other posts on this label.




























Answer: $z=\pm (a+ib), \pm (a-ib)$

Complex reducible to quadratics equation: Solve $$z^4+16z^2+100=0$$ .

Question: Solve $$z^4+16z^2+100=0$$ .

Show the answers are $$z=\pm(1+3i), \pm(1-3i)\ \ .$$

Complex reducible to quadratic equation: Solve $$z^4+12z^2+169=0$$

 



































Try solving this one now

$$z^4+16z^2+100=0$$ and show the solutions are ....https://dipa2023t2.blogspot.com/2023/06/complex-reducible-to-quadratics.html

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